#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
typedef pair<int, int> pii;
typedef pair<int, pii> pipii;
typedef pair<pii, int> piipi;
typedef pair<pii, pii> piipii;

#define mp make_pair
#define fi first
#define se second
#define all(a) (a).begin(), (a).end()
#define sz(a) (int)(a).size()
#define eb emplace_back
const int mod = 1000000007;

int l[3005], r[3005], mx[3005];
char s[3005];
int sum[3005][2];
int dp[3005][3005];
int main(){
    int n, m;
    scanf("%d%d", &n, &m);   
    scanf("%s", s+1);
    for(int i=1;i<=n;i++) mx[i] = i;
    for(int i=1;i<=m;i++){
        scanf("%d%d", &l[i], &r[i]);
        mx[l[i]] = max(mx[l[i]], r[i]);
    }

    for(int i=1;i<=n;i++){
        sum[i][s[i]-'0']++;
        sum[i][0] += sum[i-1][0];
        sum[i][1] += sum[i-1][1];
    }

    for(int i=1;i<=n;i++) mx[i] = max(mx[i], mx[i-1]);

    dp[0][0] = 1;
    for(int i=1;i<=n;i++){
        for(int j=0;j<=n;j++){
            if(dp[i-1][j]){
                int c0 = sum[mx[i]][0] - j;
                int c1 = sum[mx[i]][1] - (i-1-j);
                if(c0 > 0) dp[i][j+1] = (dp[i][j+1] + dp[i-1][j])%mod;
                if(c1 > 0) dp[i][j] = (dp[i][j] + dp[i-1][j])%mod;
            }
        }
    }
    int ans = 0;
    for(int j=0;j<=n;j++) ans = (ans + dp[n][j])%mod;
    printf("%d\n", ans);
}
### 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;

typedef long long ll;
typedef pair<int, int> pii;
typedef pair<int, pii> pipii;
typedef pair<pii, int> piipi;
typedef pair<pii, pii> piipii;

#define mp make_pair
#define fi first
#define se second
#define all(a) (a).begin(), (a).end()
#define sz(a) (int)(a).size()
#define eb emplace_back
const int mod = 1000000007;

int l[3005], r[3005], mx[3005];
char s[3005];
int sum[3005][2];
int dp[3005][3005];
int main(){
    int n, m;
    scanf("%d%d", &n, &m);   
    scanf("%s", s+1);
    for(int i=1;i<=n;i++) mx[i] = i;
    for(int i=1;i<=m;i++){
        scanf("%d%d", &l[i], &r[i]);
        mx[l[i]] = max(mx[l[i]], r[i]);
    }

    for(int i=1;i<=n;i++){
        sum[i][s[i]-'0']++;
        sum[i][0] += sum[i-1][0];
        sum[i][1] += sum[i-1][1];
    }

    for(int i=1;i<=n;i++) mx[i] = max(mx[i], mx[i-1]);

    dp[0][0] = 1;
    for(int i=1;i<=n;i++){
        for(int j=0;j<=n;j++){
            if(dp[i-1][j]){
                int c0 = sum[mx[i]][0] - j;
                int c1 = sum[mx[i]][1] - (i-1-j);
                if(c0 > 0) dp[i][j+1] = (dp[i][j+1] + dp[i-1][j])%mod;
                if(c1 > 0) dp[i][j] = (dp[i][j] + dp[i-1][j])%mod;
            }
        }
    }
    int ans = 0;
    for(int j=0;j<=n;j++) ans = (ans + dp[n][j])%mod;
    printf("%d\n", ans);
}
```