CodeForces

Levko loves array a₁, a₂, ... , a_n, consisting of integers, very much. That is why Levko is playing with array a, performing all sorts of operations with it. Each operation Levko performs is of one of two types:

1. Increase all elements from l_i to r_i by d_i. In other words, perform assignments a_j = a_j + d_i for all j that meet the inequation l_i ≤ j ≤ r_i.
2. Find the maximum of elements from l_i to r_i. That is, calculate the value .

Sadly, Levko has recently lost his array. Fortunately, Levko has records of all operations he has performed on array a. Help Levko, given the operation records, find at least one suitable array. The results of all operations for the given array must coincide with the record results. Levko clearly remembers that all numbers in his array didn't exceed 10⁹ in their absolute value, so he asks you to find such an array.

Input

The first line contains two integers n and m (1 ≤ n, m ≤ 5000) — the size of the array and the number of operations in Levko's records, correspondingly.

Next m lines describe the operations, the i-th line describes thei-th operation. The first integer in the i-th line is integer t_i(1 ≤ t_i ≤ 2) that describes the operation type. If t_i = 1, then it is followed by three integers l_i, r_i and d_i (1 ≤ l_i ≤ r_i ≤ n,  - 10⁴ ≤ d_i ≤ 10⁴) — the description of the operation of the first type. If t_i = 2, then it is followed by three integers l_i, r_i andm_i (1 ≤ l_i ≤ r_i ≤ n,  - 5·10⁷ ≤ m_i ≤ 5·10⁷) — the description of the operation of the second type.

The operations are given in the order Levko performed them on his array.

Output

In the first line print "YES" (without the quotes), if the solution exists and "NO" (without the quotes) otherwise.

If the solution exists, then on the second line print n integersa₁, a₂, ... , a_n (|a_i| ≤ 10⁹) — the recovered array.

Example

Input

4 51 2 3 12 1 2 82 3 4 71 1 3 32 3 4 8

Output

YES4 7 4 7

Input

4 51 2 3 12 1 2 82 3 4 71 1 3 32 3 4 13

Output

NO

思路: 这个题其实想好了挺好实现的,我们可以将所有的数据初始化为inf（10^9），

先从尾往后跑判断每个位置可能存在的最大值,然后在从后往前跑判断是否满足条件

#include<stdio.h>#include<math.h>#include<string.h>#define inf 1000000000using namespace std;struct node{int l,r,f,w;}q[5555];int n,m,ans1[5555],ans2[5555];int main(){scanf("%d%d",&n,&m);for(int i=1;i<=m;i++)scanf("%d%d%d%d",&q[i].f,&q[i].l,&q[i].r,&q[i].w);for(int i=1;i<=n;i++)ans1[i]=inf;for(int i=m;i>=1;i--){if(q[i].f==1){for(int j=q[i].l;j<=q[i].r;j++){if(ans1[j]!=inf)ans1[j]-=q[i].w;}}else{for(int j=q[i].l;j<=q[i].r;j++){if(ans1[j]>q[i].w)ans1[j]=q[i].w;}}}for(int i=1;i<=n;i++)ans2[i]=ans1[i];int flag=1;for(int i=1;i<=m;i++){if(q[i].f==1){for(int j=q[i].l;j<=q[i].r;j++){ans1[j]+=q[i].w;}}else{int maxn=-inf;for(int j=q[i].l;j<=q[i].r;j++){if(ans1[j]>maxn)maxn=ans1[j];}if(maxn!=q[i].w){flag=0;break;}}}if(!flag)printf("NO\n");else{printf("YES\n");printf("%d",ans2[1]);for(int j=2;j<=n;j++)printf(" %d",ans2[j]);printf("\n");}return 0;}