#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 5;
const int SN = N * 4;
const int lim = 1e9;
struct data {
  int ans[5][5];
};
data segtree[SN];
char s[N];
int n, q;
int l, r;
data mz(char c) {
  data res;
  for (int i = 0; i < 5; ++i) {
    for (int j = 0; j < 5; ++j) {
      res.ans[i][j] = i == j ? 0 : lim;
    }
  }
  if (c == '2') {
    res.ans[0][1] = 0;
    res.ans[0][0] = 1;
  }
  if (c == '0') {
    res.ans[1][2] = 0;
    res.ans[1][1] = 1;
  }
  if (c == '1') {
    res.ans[2][3] = 0;
    res.ans[2][2] = 1;
  }
  if (c == '7') {
    res.ans[3][4] = 0;
    res.ans[3][3] = 1;
  }
  if (c == '6') {
    res.ans[3][3] = 1;
    res.ans[4][4] = 1;
  }
  return res;
}
data Merger(data l, data r) {
  data res;
  for (int i = 0; i < 5; ++i) {
    for (int j = 0; j < 5; ++j) {
      res.ans[i][j] = lim;
      for (int k = 0; k < 5; ++k) {
        res.ans[i][j] = min(res.ans[i][j], l.ans[i][k] + r.ans[k][j]);
      }
    }
  }
  return res;
}
void build(int node, int u, int v) {
  if (u == v) {
    segtree[node] = mz(s[u]);
    return;
  }
  int m = (u + v) >> 1;
  build(node << 1, u, m);
  build((node << 1) | 1, m + 1, v);
  segtree[node] = Merger(segtree[node << 1], segtree[(node << 1) | 1]);
}
data query(int node, int u, int v, int l, int r) {
  if (u > v | u > r | v < l) {
    return mz('$');
  }
  if (u >= l && v <= r) {
    return segtree[node];
  }
  int m = (u + v) >> 1;
  return Merger(query(node * 2, u, m, l, r),
                query(1 + (node * 2), m + 1, v, l, r));
}
int main() {
  scanf("%d %d %s", &n, &q, s);
  build(1, 0, n - 1);
  while (q-- > 0) {
    scanf("%d %d", &l, &r);
    data res = query(1, 0, n - 1, l - 1, r - 1);
    int answer = res.ans[0][4] == lim ? -1 : res.ans[0][4];
    printf("%d\n", answer);
  }
  return 0;
}

### Prompt

Your task is to create a CPP solution 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 = 2e5 + 5;
const int SN = N * 4;
const int lim = 1e9;
struct data {
  int ans[5][5];
};
data segtree[SN];
char s[N];
int n, q;
int l, r;
data mz(char c) {
  data res;
  for (int i = 0; i < 5; ++i) {
    for (int j = 0; j < 5; ++j) {
      res.ans[i][j] = i == j ? 0 : lim;
    }
  }
  if (c == '2') {
    res.ans[0][1] = 0;
    res.ans[0][0] = 1;
  }
  if (c == '0') {
    res.ans[1][2] = 0;
    res.ans[1][1] = 1;
  }
  if (c == '1') {
    res.ans[2][3] = 0;
    res.ans[2][2] = 1;
  }
  if (c == '7') {
    res.ans[3][4] = 0;
    res.ans[3][3] = 1;
  }
  if (c == '6') {
    res.ans[3][3] = 1;
    res.ans[4][4] = 1;
  }
  return res;
}
data Merger(data l, data r) {
  data res;
  for (int i = 0; i < 5; ++i) {
    for (int j = 0; j < 5; ++j) {
      res.ans[i][j] = lim;
      for (int k = 0; k < 5; ++k) {
        res.ans[i][j] = min(res.ans[i][j], l.ans[i][k] + r.ans[k][j]);
      }
    }
  }
  return res;
}
void build(int node, int u, int v) {
  if (u == v) {
    segtree[node] = mz(s[u]);
    return;
  }
  int m = (u + v) >> 1;
  build(node << 1, u, m);
  build((node << 1) | 1, m + 1, v);
  segtree[node] = Merger(segtree[node << 1], segtree[(node << 1) | 1]);
}
data query(int node, int u, int v, int l, int r) {
  if (u > v | u > r | v < l) {
    return mz('$');
  }
  if (u >= l && v <= r) {
    return segtree[node];
  }
  int m = (u + v) >> 1;
  return Merger(query(node * 2, u, m, l, r),
                query(1 + (node * 2), m + 1, v, l, r));
}
int main() {
  scanf("%d %d %s", &n, &q, s);
  build(1, 0, n - 1);
  while (q-- > 0) {
    scanf("%d %d", &l, &r);
    data res = query(1, 0, n - 1, l - 1, r - 1);
    int answer = res.ans[0][4] == lim ? -1 : res.ans[0][4];
    printf("%d\n", answer);
  }
  return 0;
}

```