codeforces_652C. Foe Pairs

C. Foe Pairs

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given a permutation p of length n. Also you are given m foe pairs (ai, bi) (1 ≤ ai, bi ≤ n, ai ≠ bi).

Your task is to count the number of different intervals (x, y) (1 ≤ x ≤ y ≤ n) that do not contain any foe pairs. So you shouldn't count intervals (x, y) that contain at least one foe pair in it (the positions and order of the values from the foe pair are not important).

Consider some example: p = [1, 3, 2, 4] and foe pairs are {(3, 2), (4, 2)}. The interval (1, 3) is incorrect because it contains a foe pair(3, 2). The interval (1, 4) is also incorrect because it contains two foe pairs (3, 2) and (4, 2). But the interval (1, 2) is correct because it doesn't contain any foe pair.

Input

The first line contains two integers n and m (1 ≤ n, m ≤ 3·105) — the length of the permutation p and the number of foe pairs.

The second line contains n distinct integers pi (1 ≤ pi ≤ n) — the elements of the permutation p.

Each of the next m lines contains two integers (ai, bi) (1 ≤ ai, bi ≤ n, ai ≠ bi) — the i-th foe pair. Note a foe pair can appear multiple times in the given list.

Output

Print the only integer c — the number of different intervals (x, y) that does not contain any foe pairs.

Note that the answer can be too large, so you should use 64-bit integer type to store it. In C++ you can use the long long integer type and in Java you can use long integer type.

Examples

input

4 21 3 2 43 22 4

output

5

input

9 59 7 2 3 1 4 6 5 81 64 52 77 22 7

output

20

用v数组保存每个数的指标，b[x]表示指标x上的数的pair对数的指标，若把x称为起点，则b[x]为终点，没有pair对数的指标指向指标n，bad[x]表示当前指标x上有几个终点。

#include <iostream>#include <cstdio>#include <cstdlib>#include <cstring>#include <cmath>#include <stack>#include <bitset>#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 inf 0x3f3f3f3f#define maxn 300010#define mod 1000000007#define PI acos(-1.0)#define LL long longusing namespace std;int a[maxn];int b[maxn];int v[maxn];int bad[maxn];int main(){    int i,n,m;    mem(b,0);    mem(bad,0);    Si(n);Si(m);    for(i=0;i<n;i++)    {        Si(a[i]);        v[a[i]]=i;        b[i]=n;    }    W(m)    {        int x,y;        Si(x);        Si(y);        x=v[x];        y=v[y];        if(x>y)swap(x,y);        b[x]=min(b[x],y);    }    int l,r=0;    LL ans=0;    for(l=0;l<n;l++)    {        while(r<n&&bad[r]==0)        {            bad[b[r]]++;            r++;        }        ans+=r-l;        bad[b[l]]--;    }    Pl(ans);    return 0;}

Note

In the first example the intervals from the answer are (1, 1), (1, 2), (2, 2), (3, 3) and (4, 4).