#include <bits/stdc++.h>

const int Mod=1e9+7,N=3e3+10;

int n,m,dp[N][N],summ[N];
char s[N];

void Init() {
	scanf("%d%d",&n,&m);
	scanf("%s",s+1);
	for (int i=1;i<=n;++i) summ[i]=summ[i-1]+(s[i]=='0'); 
}

void Solve() {
	int nowr=0,l,r,cnt=0;
	scanf("%d%d",&l,&r); ++cnt;
	dp[0][0]=1;
	for (int i=0;i<n;++i) {
		int sum=0; 
		while (i+1==l && cnt<=m) {
			if (r>nowr) sum+=summ[r]-summ[nowr],nowr=r;
			if (cnt<m) { scanf("%d%d",&l,&r); ++cnt; }
			else break; 
		}
		if (nowr>i){
			for (int j=0;j<=nowr-i;++j) {
				if (j+sum) dp[i+1][j+sum-1]=(dp[i+1][j+sum-1]+dp[i][j])%Mod;
				if (nowr-i-j-sum) dp[i+1][j+sum]=(dp[i+1][j+sum]+dp[i][j])%Mod; 
			}
		}else dp[i+1][0]=dp[i][0],nowr=i+1;
	}
	printf("%d\n",dp[n][0]);
} 

int main() {
	Init();
	Solve();
	return 0;
}
### Prompt

Please provide a Cpp coded solution to the problem described below:

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>

const int Mod=1e9+7,N=3e3+10;

int n,m,dp[N][N],summ[N];
char s[N];

void Init() {
	scanf("%d%d",&n,&m);
	scanf("%s",s+1);
	for (int i=1;i<=n;++i) summ[i]=summ[i-1]+(s[i]=='0'); 
}

void Solve() {
	int nowr=0,l,r,cnt=0;
	scanf("%d%d",&l,&r); ++cnt;
	dp[0][0]=1;
	for (int i=0;i<n;++i) {
		int sum=0; 
		while (i+1==l && cnt<=m) {
			if (r>nowr) sum+=summ[r]-summ[nowr],nowr=r;
			if (cnt<m) { scanf("%d%d",&l,&r); ++cnt; }
			else break; 
		}
		if (nowr>i){
			for (int j=0;j<=nowr-i;++j) {
				if (j+sum) dp[i+1][j+sum-1]=(dp[i+1][j+sum-1]+dp[i][j])%Mod;
				if (nowr-i-j-sum) dp[i+1][j+sum]=(dp[i+1][j+sum]+dp[i][j])%Mod; 
			}
		}else dp[i+1][0]=dp[i][0],nowr=i+1;
	}
	printf("%d\n",dp[n][0]);
} 

int main() {
	Init();
	Solve();
	return 0;
}
```