#include <bits/stdc++.h>
using namespace std;
const int N = 1 << 18;
struct Tuple {
  int dp[5][5];
  Tuple(char c) : Tuple() {
    if (c == '2') {
      dp[0][0] = 1;
      dp[0][1] = 0;
    } else if (c == '0') {
      dp[1][1] = 1;
      dp[1][2] = 0;
    } else if (c == '1') {
      dp[2][2] = 1;
      dp[2][3] = 0;
    } else if (c == '6') {
      dp[3][3] = 1;
      dp[4][4] = 1;
    } else if (c == '7') {
      dp[3][3] = 1;
      dp[3][4] = 0;
    }
  }
  Tuple() {
    for (int i = 0; i < 5; i++) {
      for (int j = 0; j < 5; j++) {
        dp[i][j] = 1e7;
      }
      dp[i][i] = 0;
    }
  }
};
Tuple dat[N * 2];
Tuple operator*(Tuple a, Tuple b) {
  Tuple ret;
  for (int i = 0; i < 5; i++) {
    ret.dp[i][i] = 1e7;
  }
  for (int i = 0; i < 5; i++) {
    for (int j = i; j < 5; j++) {
      for (int k = j; k < 5; k++) {
        ret.dp[i][k] = min(ret.dp[i][k], a.dp[i][j] + b.dp[j][k]);
      }
    }
  }
  return ret;
}
Tuple query(int l, int r) {
  Tuple vl, vr;
  l += N;
  r += N;
  for (; l < r; l /= 2, r /= 2) {
    if (l & 1) vl = vl * dat[l++];
    if (r & 1) vr = dat[--r] * vr;
  }
  return vl * vr;
}
int main() {
  int n, q;
  cin >> n >> q;
  string s;
  cin >> s;
  for (int i = 0; i < n; i++) {
    dat[i + N] = Tuple(s[i]);
  }
  for (int i = N - 1; i >= 1; i--) {
    dat[i] = dat[i * 2] * dat[i * 2 + 1];
  }
  for (int i = 0; i < q; i++) {
    int l, r;
    scanf("%d %d", &l, &r);
    Tuple ret = query(l - 1, r);
    int ans = ret.dp[0][4];
    if (ans > n) ans = -1;
    printf("%d\n", ans);
  }
}

### Prompt

Please create a solution in cpp to the following problem:

A string t is called nice if a string "2017" occurs in t as a subsequence but a string "2016" doesn't occur in t as a subsequence. For example, strings "203434107" and "9220617" are nice, while strings "20016", "1234" and "20167" aren't nice.

The ugliness of a string is the minimum possible number of characters to remove, in order to obtain a nice string. If it's impossible to make a string nice by removing characters, its ugliness is  - 1.

Limak has a string s of length n, with characters indexed 1 through n. He asks you q queries. In the i-th query you should compute and print the ugliness of a substring (continuous subsequence) of s starting at the index ai and ending at the index bi (inclusive).

Input

The first line of the input contains two integers n and q (4 ≤ n ≤ 200 000, 1 ≤ q ≤ 200 000) — the length of the string s and the number of queries respectively.

The second line contains a string s of length n. Every character is one of digits '0'–'9'.

The i-th of next q lines contains two integers ai and bi (1 ≤ ai ≤ bi ≤ n), describing a substring in the i-th query.

Output

For each query print the ugliness of the given substring.

Examples

Input

8 3
20166766
1 8
1 7
2 8


Output

4
3
-1


Input

15 5
012016662091670
3 4
1 14
4 15
1 13
10 15


Output

-1
2
1
-1
-1


Input

4 2
1234
2 4
1 2


Output

-1
-1

Note

In the first sample:

  * In the first query, ugliness("20166766") = 4 because all four sixes must be removed. 
  * In the second query, ugliness("2016676") = 3 because all three sixes must be removed. 
  * In the third query, ugliness("0166766") = - 1 because it's impossible to remove some digits to get a nice string. 



In the second sample:

  * In the second query, ugliness("01201666209167") = 2. It's optimal to remove the first digit '2' and the last digit '6', what gives a string "010166620917", which is nice. 
  * In the third query, ugliness("016662091670") = 1. It's optimal to remove the last digit '6', what gives a nice string "01666209170". 

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 1 << 18;
struct Tuple {
  int dp[5][5];
  Tuple(char c) : Tuple() {
    if (c == '2') {
      dp[0][0] = 1;
      dp[0][1] = 0;
    } else if (c == '0') {
      dp[1][1] = 1;
      dp[1][2] = 0;
    } else if (c == '1') {
      dp[2][2] = 1;
      dp[2][3] = 0;
    } else if (c == '6') {
      dp[3][3] = 1;
      dp[4][4] = 1;
    } else if (c == '7') {
      dp[3][3] = 1;
      dp[3][4] = 0;
    }
  }
  Tuple() {
    for (int i = 0; i < 5; i++) {
      for (int j = 0; j < 5; j++) {
        dp[i][j] = 1e7;
      }
      dp[i][i] = 0;
    }
  }
};
Tuple dat[N * 2];
Tuple operator*(Tuple a, Tuple b) {
  Tuple ret;
  for (int i = 0; i < 5; i++) {
    ret.dp[i][i] = 1e7;
  }
  for (int i = 0; i < 5; i++) {
    for (int j = i; j < 5; j++) {
      for (int k = j; k < 5; k++) {
        ret.dp[i][k] = min(ret.dp[i][k], a.dp[i][j] + b.dp[j][k]);
      }
    }
  }
  return ret;
}
Tuple query(int l, int r) {
  Tuple vl, vr;
  l += N;
  r += N;
  for (; l < r; l /= 2, r /= 2) {
    if (l & 1) vl = vl * dat[l++];
    if (r & 1) vr = dat[--r] * vr;
  }
  return vl * vr;
}
int main() {
  int n, q;
  cin >> n >> q;
  string s;
  cin >> s;
  for (int i = 0; i < n; i++) {
    dat[i + N] = Tuple(s[i]);
  }
  for (int i = N - 1; i >= 1; i--) {
    dat[i] = dat[i * 2] * dat[i * 2 + 1];
  }
  for (int i = 0; i < q; i++) {
    int l, r;
    scanf("%d %d", &l, &r);
    Tuple ret = query(l - 1, r);
    int ans = ret.dp[0][4];
    if (ans > n) ans = -1;
    printf("%d\n", ans);
  }
}

```