#include <bits/stdc++.h>
using namespace std;
const int kN = 200010, kM = 5;
int n, q;
char s[kN];
struct State {
  int m[kM][kM];
  State() {
    for (int i = 0; i < kM; ++i) {
      for (int j = 0; j < kM; ++j) m[i][j] = kN;
    }
  }
  State operator*(const State& b) const {
    State a(*this), res;
    for (int i = 0; i < kM; ++i) {
      for (int j = 0; j < kM; ++j) {
        for (int k = 0; k < kM; ++k)
          res.m[i][j] = min(res.m[i][j], a.m[i][k] + b.m[k][j]);
      }
    }
    return res;
  }
} tree[kN << 2];
void build(int l, int r, int root) {
  if (l == r) {
    tree[root].m[0][0] = 0;
    tree[root].m[1][1] = 0;
    tree[root].m[2][2] = 0;
    tree[root].m[3][3] = 0;
    tree[root].m[4][4] = 0;
    if (s[l] == '2') {
      tree[root].m[0][0] = 1;
      tree[root].m[0][1] = 0;
    } else if (s[l] == '0') {
      tree[root].m[1][1] = 1;
      tree[root].m[1][2] = 0;
    } else if (s[l] == '1') {
      tree[root].m[2][2] = 1;
      tree[root].m[2][3] = 0;
    } else if (s[l] == '7') {
      tree[root].m[3][3] = 1;
      tree[root].m[3][4] = 0;
    } else if (s[l] == '6') {
      tree[root].m[3][3] = 1;
      tree[root].m[4][4] = 1;
    }
    return;
  }
  int mid = (l + r) >> 1;
  build(l, mid, root << 1);
  build(mid + 1, r, root << 1 | 1);
  tree[root] = tree[root << 1] * tree[root << 1 | 1];
}
State query(int L, int R, int l, int r, int root) {
  if (L == l && r == R) {
    return tree[root];
  }
  int mid = (l + r) >> 1;
  if (R <= mid)
    return query(L, R, l, mid, root << 1);
  else if (L > mid)
    return query(L, R, mid + 1, r, root << 1 | 1);
  else
    return query(L, mid, l, mid, root << 1) *
           query(mid + 1, R, mid + 1, r, root << 1 | 1);
}
void getData() { scanf("%d%d%s", &n, &q, (s + 1)); }
void solve() {
  int l, r;
  build(1, n, 1);
  while (q--) {
    scanf("%d%d", &l, &r);
    int ans = query(l, r, 1, n, 1).m[0][4];
    if (ans == kN) ans = -1;
    printf("%d\n", ans);
  }
}
int main() {
  getData();
  solve();
  return 0;
}

### 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 kN = 200010, kM = 5;
int n, q;
char s[kN];
struct State {
  int m[kM][kM];
  State() {
    for (int i = 0; i < kM; ++i) {
      for (int j = 0; j < kM; ++j) m[i][j] = kN;
    }
  }
  State operator*(const State& b) const {
    State a(*this), res;
    for (int i = 0; i < kM; ++i) {
      for (int j = 0; j < kM; ++j) {
        for (int k = 0; k < kM; ++k)
          res.m[i][j] = min(res.m[i][j], a.m[i][k] + b.m[k][j]);
      }
    }
    return res;
  }
} tree[kN << 2];
void build(int l, int r, int root) {
  if (l == r) {
    tree[root].m[0][0] = 0;
    tree[root].m[1][1] = 0;
    tree[root].m[2][2] = 0;
    tree[root].m[3][3] = 0;
    tree[root].m[4][4] = 0;
    if (s[l] == '2') {
      tree[root].m[0][0] = 1;
      tree[root].m[0][1] = 0;
    } else if (s[l] == '0') {
      tree[root].m[1][1] = 1;
      tree[root].m[1][2] = 0;
    } else if (s[l] == '1') {
      tree[root].m[2][2] = 1;
      tree[root].m[2][3] = 0;
    } else if (s[l] == '7') {
      tree[root].m[3][3] = 1;
      tree[root].m[3][4] = 0;
    } else if (s[l] == '6') {
      tree[root].m[3][3] = 1;
      tree[root].m[4][4] = 1;
    }
    return;
  }
  int mid = (l + r) >> 1;
  build(l, mid, root << 1);
  build(mid + 1, r, root << 1 | 1);
  tree[root] = tree[root << 1] * tree[root << 1 | 1];
}
State query(int L, int R, int l, int r, int root) {
  if (L == l && r == R) {
    return tree[root];
  }
  int mid = (l + r) >> 1;
  if (R <= mid)
    return query(L, R, l, mid, root << 1);
  else if (L > mid)
    return query(L, R, mid + 1, r, root << 1 | 1);
  else
    return query(L, mid, l, mid, root << 1) *
           query(mid + 1, R, mid + 1, r, root << 1 | 1);
}
void getData() { scanf("%d%d%s", &n, &q, (s + 1)); }
void solve() {
  int l, r;
  build(1, n, 1);
  while (q--) {
    scanf("%d%d", &l, &r);
    int ans = query(l, r, 1, n, 1).m[0][4];
    if (ans == kN) ans = -1;
    printf("%d\n", ans);
  }
}
int main() {
  getData();
  solve();
  return 0;
}

```