#include <bits/stdc++.h>
using namespace std;
const int N=3005,Mod=1e9+7;
int dp[N][N];//dp[i][j]表示前i个位置填满了，后面有j个1可以随便填的方案数 
int r[N],a[N],sum[N];
char s[N];
inline int Max(int a,int b){
	return a>b?a:b;
}
inline int Min(int a,int b){
	return a<b?a:b;
}
inline void mo(int&x){
	x=(x<Mod?x:x-Mod);
}
int main (){
	int n,m;scanf ("%d%d",&n,&m);
	scanf ("%s",s+1);
	for (int i=1;i<=n;i++) a[i]=s[i]-'0';
	for (int i=1;i<=n+1;i++) r[i]=i,sum[i]=sum[i-1]+a[i];
	for (int i=1,x,y;i<=m;i++){
		scanf ("%d%d",&x,&y);
		r[x]=Max(r[x],y);
	}
	for (int i=1;i<=n;i++) r[i]=Max(r[i],r[i-1]);
	dp[1][sum[r[1]]]=1;
	for (int i=1;i<=n;i++)
		for (int j=0;j<=n;j++)
			if (dp[i][j]){
				int L=r[i]+1,R=r[i+1];
				if (j) mo(dp[i+1][j-1-sum[L-1]+sum[R]]+=dp[i][j]);
				if (r[i]-i-j+1) mo(dp[i+1][j-sum[L-1]+sum[R]]+=dp[i][j]);
			}
	printf ("%d",dp[n+1][0]);
	return 0;
}

### Prompt

Create a solution in cpp for the following problem:

There is a string S of length N consisting of characters `0` and `1`. You will perform the following operation for each i = 1, 2, ..., m:

* Arbitrarily permute the characters within the substring of S starting at the l_i-th character from the left and extending through the r_i-th character.



Here, the sequence l_i is non-decreasing.

How many values are possible for S after the M operations, modulo 1000000007(= 10^9+7)?

Constraints

* 2≦N≦3000
* 1≦M≦3000
* S consists of characters `0` and `1`.
* The length of S equals N.
* 1≦l_i < r_i≦N
* l_i ≦ l_{i+1}

Input

The input is given from Standard Input in the following format:


N M
S
l_1 r_1
:
l_M r_M


Output

Print the number of the possible values for S after the M operations, modulo 1000000007.

Examples

Input

5 2
01001
2 4
3 5


Output

6


Input

9 3
110111110
1 4
4 6
6 9


Output

26


Input

11 6
00101000110
2 4
2 3
4 7
5 6
6 10
10 11


Output

143

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int N=3005,Mod=1e9+7;
int dp[N][N];//dp[i][j]表示前i个位置填满了，后面有j个1可以随便填的方案数 
int r[N],a[N],sum[N];
char s[N];
inline int Max(int a,int b){
	return a>b?a:b;
}
inline int Min(int a,int b){
	return a<b?a:b;
}
inline void mo(int&x){
	x=(x<Mod?x:x-Mod);
}
int main (){
	int n,m;scanf ("%d%d",&n,&m);
	scanf ("%s",s+1);
	for (int i=1;i<=n;i++) a[i]=s[i]-'0';
	for (int i=1;i<=n+1;i++) r[i]=i,sum[i]=sum[i-1]+a[i];
	for (int i=1,x,y;i<=m;i++){
		scanf ("%d%d",&x,&y);
		r[x]=Max(r[x],y);
	}
	for (int i=1;i<=n;i++) r[i]=Max(r[i],r[i-1]);
	dp[1][sum[r[1]]]=1;
	for (int i=1;i<=n;i++)
		for (int j=0;j<=n;j++)
			if (dp[i][j]){
				int L=r[i]+1,R=r[i+1];
				if (j) mo(dp[i+1][j-1-sum[L-1]+sum[R]]+=dp[i][j]);
				if (r[i]-i-j+1) mo(dp[i+1][j-sum[L-1]+sum[R]]+=dp[i][j]);
			}
	printf ("%d",dp[n+1][0]);
	return 0;
}

```