codeforces_607A. Chain Reaction（二分+简单dp）

A. Chain Reaction

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

There are n beacons located at distinct positions on a number line. The i-th beacon has position ai and power level bi. When the i-th beacon is activated, it destroys all beacons to its left (direction of decreasing coordinates) within distance bi inclusive. The beacon itself is not destroyed however. Saitama will activate the beacons one at a time from right to left. If a beacon is destroyed, it cannot be activated.

Saitama wants Genos to add a beacon strictly to the right of all the existing beacons, with any position and any power level, such that the least possible number of beacons are destroyed. Note that Genos's placement of the beacon means it will be the first beacon activated. Help Genos by finding the minimum number of beacons that could be destroyed.

Input

The first line of input contains a single integer n (1 ≤ n ≤ 100 000) — the initial number of beacons.

The i-th of next n lines contains two integers ai and bi (0 ≤ ai ≤ 1 000 000, 1 ≤ bi ≤ 1 000 000) — the position and power level of thei-th beacon respectively. No two beacons will have the same position, so ai ≠ aj if i ≠ j.

Output

Print a single integer — the minimum number of beacons that could be destroyed if exactly one beacon is added.

Examples

input

41 93 16 17 4

output

1

input

71 12 13 14 15 16 17 1

output

3

Note

For the first sample case, the minimum number of beacons destroyed is 1. One way to achieve this is to place a beacon at position 9 with power level 2.

For the second sample case, the minimum number of beacons destroyed is 3. One way to achieve this is to place a beacon at position1337 with power level 42.

二分找出每个灯塔能破坏几个已存在的灯塔，然后dp即可。不过我卡在了奇怪的地方。。一开始pos<0的情况dp居然没有置1.。。。吃个教训

#include <iostream>#include <cstdio>#include <cstdlib>#include <cstring>#include <cmath>#include <stack>#include <bitset>#include <queue>#include <set>#include <map>#include <string>#include <algorithm>#define Si(a) scanf("%d",&a)#define Sl(a) scanf("%lld",&a)#define Sd(a) scanf("%lf",&a)#define Ss(a) scanf("%s",a)#define Pi(a) printf("%d\n",(a))#define Pl(a) printf("%lld\n",(a))#define Pd(a) printf("%lf\n",(a))#define Ps(a) printf("%s\n",(a))#define W(a) while(a--)#define mem(a,b) memset(a,(b),sizeof(a))#define FOP freopen("data.txt","r",stdin)#define inf 0x3f3f3f3f#define maxn 100010#define mod 1000000007#define PI acos(-1.0)#define LL long longusing namespace std;struct node{    int x;    int v;    int kn;}a[maxn];bool cmp(node a,node b){    return a.x<b.x;}int dp[maxn];int l_search(int size, int key){    int first = 0, middle;    int half, len;    len = size;    while(len > 0) {        half = len >> 1;        middle = first + half;        if(a[middle].x < key) {            first = middle + 1;            len = len-half-1;        }        else            len = half;    }    return first;}int main(){    //FOP;    int i,n;    Si(n);    for(i=0;i<n;i++)    {        Si(a[i].x),Si(a[i].v);    }    sort(a,a+n,cmp);    a[0].kn=0;    int key,pos,num;    for(i=1;i<n;i++)    {        key=a[i].x-a[i].v;        pos=l_search(i,key);        num=i-pos;        a[i].kn=num;    }    dp[0]=1;    for(i=1;i<n;i++)    {        pos=i-a[i].kn-1;        if(pos<0)dp[i]=1;        dp[i]=dp[pos]+1;    }    int ma=0;    for(i=0;i<n;i++)    {        if(ma<dp[i])ma=dp[i];    }    Pi(n-ma);    return 0;}