#include <bits/stdc++.h>
using namespace std;
struct sub {
  int dp[5][5];
  sub() {
    for (int i = 0; i < 5; i++) {
      for (int j = 0; j < 5; j++) {
        dp[i][j] = 200000;
      }
      dp[i][i] = 0;
    }
  }
  sub operator+(const sub &q) const {
    sub r;
    for (int i = 0; i < 5; i++) r.dp[i][i] = 200000;
    for (int i = 0; i < 5; i++) {
      for (int j = i; j < 5; j++) {
        for (int k = j; k < 5; k++) {
          r.dp[i][k] = min(r.dp[i][k], dp[i][j] + q.dp[j][k]);
        }
      }
    }
    return r;
  }
};
int n, q;
sub seg[1 << 19];
char s[200005];
void in(int l, int r, int k) {
  if (l == r) {
    int o = k - (1 << 18) + 1;
    if (o >= n) return;
    if (s[o] == '2') {
      seg[k].dp[0][1] = 0;
      seg[k].dp[0][0] = 1;
    } else if (s[o] == '0') {
      seg[k].dp[1][2] = 0;
      seg[k].dp[1][1] = 1;
    } else if (s[o] == '1') {
      seg[k].dp[2][3] = 0;
      seg[k].dp[2][2] = 1;
    } else if (s[o] == '7') {
      seg[k].dp[3][4] = 0;
      seg[k].dp[3][3] = 1;
    } else if (s[o] == '6') {
      seg[k].dp[3][3] = 1;
      seg[k].dp[4][4] = 1;
    }
  } else {
    in(l, (r + l - 1) / 2, k * 2 + 1);
    in((l + r + 1) / 2, r, k * 2 + 2);
    seg[k] = seg[k * 2 + 1] + seg[k * 2 + 2];
  }
}
sub que(int a, int b, int l, int r, int k) {
  if (b < l || r < a) {
    sub r;
    return r;
  }
  if (a <= l && r <= b) return seg[k];
  return que(a, b, l, (l + r - 1) / 2, k * 2 + 1) +
         que(a, b, (l + r + 1) / 2, r, k * 2 + 2);
}
int main(void) {
  cin >> n >> q >> s;
  in(0, (1 << 18) - 1, 0);
  for (int i = 0; i < q; i++) {
    int a, b;
    sub p;
    cin >> a >> b;
    a--;
    b--;
    p = que(a, b, 0, (1 << 18) - 1, 0);
    if (p.dp[0][4] == 200000)
      cout << -1 << endl;
    else
      cout << p.dp[0][4] << endl;
  }
}

### Prompt

Develop a solution in CPP to the problem described below:

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;
struct sub {
  int dp[5][5];
  sub() {
    for (int i = 0; i < 5; i++) {
      for (int j = 0; j < 5; j++) {
        dp[i][j] = 200000;
      }
      dp[i][i] = 0;
    }
  }
  sub operator+(const sub &q) const {
    sub r;
    for (int i = 0; i < 5; i++) r.dp[i][i] = 200000;
    for (int i = 0; i < 5; i++) {
      for (int j = i; j < 5; j++) {
        for (int k = j; k < 5; k++) {
          r.dp[i][k] = min(r.dp[i][k], dp[i][j] + q.dp[j][k]);
        }
      }
    }
    return r;
  }
};
int n, q;
sub seg[1 << 19];
char s[200005];
void in(int l, int r, int k) {
  if (l == r) {
    int o = k - (1 << 18) + 1;
    if (o >= n) return;
    if (s[o] == '2') {
      seg[k].dp[0][1] = 0;
      seg[k].dp[0][0] = 1;
    } else if (s[o] == '0') {
      seg[k].dp[1][2] = 0;
      seg[k].dp[1][1] = 1;
    } else if (s[o] == '1') {
      seg[k].dp[2][3] = 0;
      seg[k].dp[2][2] = 1;
    } else if (s[o] == '7') {
      seg[k].dp[3][4] = 0;
      seg[k].dp[3][3] = 1;
    } else if (s[o] == '6') {
      seg[k].dp[3][3] = 1;
      seg[k].dp[4][4] = 1;
    }
  } else {
    in(l, (r + l - 1) / 2, k * 2 + 1);
    in((l + r + 1) / 2, r, k * 2 + 2);
    seg[k] = seg[k * 2 + 1] + seg[k * 2 + 2];
  }
}
sub que(int a, int b, int l, int r, int k) {
  if (b < l || r < a) {
    sub r;
    return r;
  }
  if (a <= l && r <= b) return seg[k];
  return que(a, b, l, (l + r - 1) / 2, k * 2 + 1) +
         que(a, b, (l + r + 1) / 2, r, k * 2 + 2);
}
int main(void) {
  cin >> n >> q >> s;
  in(0, (1 << 18) - 1, 0);
  for (int i = 0; i < q; i++) {
    int a, b;
    sub p;
    cin >> a >> b;
    a--;
    b--;
    p = que(a, b, 0, (1 << 18) - 1, 0);
    if (p.dp[0][4] == 200000)
      cout << -1 << endl;
    else
      cout << p.dp[0][4] << endl;
  }
}

```