Greg has a pad. The pad's screen is an n × m rectangle, each cell can be either black or white. We'll consider the pad rows to be numbered with integers from 1 to n from top to bottom. Similarly, the pad's columns are numbered with integers from 1 to m from left to right.

Greg thinks that the pad's screen displays a cave if the following conditions hold:

  * There is a segment [l, r] (1 ≤ l ≤ r ≤ n), such that each of the rows l, l + 1, ..., r has exactly two black cells and all other rows have only white cells. 
  * There is a row number t (l ≤ t ≤ r), such that for all pairs of rows with numbers i and j (l ≤ i ≤ j ≤ t) the set of columns between the black cells in row i (with the columns where is these black cells) is the subset of the set of columns between the black cells in row j (with the columns where is these black cells). Similarly, for all pairs of rows with numbers i and j (t ≤ i ≤ j ≤ r) the set of columns between the black cells in row j (with the columns where is these black cells) is the subset of the set of columns between the black cells in row i (with the columns where is these black cells). 



Greg wondered, how many ways there are to paint a cave on his pad. Two ways can be considered distinct if there is a cell that has distinct colors on the two pictures.

Help Greg.

Input

The first line contains two integers n, m — the pad's screen size (1 ≤ n, m ≤ 2000).

Output

In the single line print the remainder after dividing the answer to the problem by 1000000007 (109 + 7).

Examples

Input

1 1


Output

0


Input

4 4


Output

485


Input

3 5


Output

451
#include<stdio.h>
#include<string.h>
#define M 100005
__int64 a[M],b[M],c[M];
typedef struct{
    int l,r;
    int d;
}queue;
queue Q[M];
int main(){
    freopen("in.txt","r",stdin);
    int n,m,k;
    int x,y,d;
    while(~scanf("%d%d%d",&n,&m,&k)){
        for(int _=1;_<=n;_++) scanf("%I64d",&a[_]);
        for(int _=1;_<=m;_++){
            scanf("%d%d%d",&Q[_].l,&Q[_].r,&Q[_].d);
        }
        for(int _=1;_<=k;_++){
            scanf("%d%d",&x,&y);
            b[x]++;b[y+1]--;
        }
        int s=0;
        for(int i=1;i<=m;i++){
            s+=b[i];c[i]+=s;
        }
        memset(b,0,sizeof(b));
        for(int i=1;i<=m;i++){
            b[Q[i].l]+=c[i]*d; b[Q[i].r+1]-=c[i]*d;
        }
        __int64 sum=0;
        for(int i=1;i<=n;i++){
            sum+=b[i]; a[i]+=sum;
        }
        for(int i=1;i<=n;i++) printf("%I64d%c",a[i],i<n?' ':'\n');
    }
    return 0;
}

		  				 	      					 				