#include <bits/stdc++.h>
using namespace std;
struct node {
  int state[5][5];
  node() {
    for (int i = 0; i < 5; i++)
      for (int j = 0; j < 5; j++) {
        state[i][j] = 1031231234;
      }
  }
  node(char v) {
    for (int i = 0; i < 5; i++)
      for (int j = 0; j < 5; j++) {
        state[i][j] = (i == j) ? 0 : 1031231234;
      }
    if (v == '2') {
      state[0][0] = 1;
      state[0][1] = 0;
    } else if (v == '0') {
      state[1][1] = 1;
      state[1][2] = 0;
    } else if (v == '1') {
      state[2][2] = 1;
      state[2][3] = 0;
    } else if (v == '6') {
      state[3][3] = 1;
      state[4][4] = 1;
    } else if (v == '7') {
      state[3][3] = 1;
      state[3][4] = 0;
    }
  }
  node combine(const node& o) const {
    node res;
    for (int i = 0; i < 5; i++) {
      for (int j = 0; j < 5; j++) {
        for (int k = 0; k < 5; k++) {
          res.state[i][j] = min(res.state[i][j], state[i][k] + o.state[k][j]);
        }
      }
    }
    return res;
  }
};
struct segmentTree {
  int sz;
  vector<node> elements;
  segmentTree(int s) {
    sz = s;
    elements.resize(2 * sz);
  }
  node& operator[](int i) { return elements[i + sz]; }
  void build() {
    for (int i = sz - 1; i >= 1; i--) {
      elements[i] = elements[i << 1].combine(elements[i << 1 | 1]);
    }
  }
  int query(int l, int r) {
    node lacc, racc;
    lacc.state[0][0] = 0, racc.state[4][4] = 0;
    l += sz;
    r += sz;
    while (l <= r) {
      if (l & 1) {
        lacc = lacc.combine(elements[l]);
        l++;
      }
      if (!(r & 1)) {
        racc = elements[r].combine(racc);
        r--;
      }
      l >>= 1;
      r >>= 1;
    }
    node res = lacc.combine(racc);
    return res.state[0][4] != 1031231234 ? res.state[0][4] : -1;
  }
};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n, q;
  cin >> n >> q;
  string s;
  cin >> s;
  segmentTree st(n);
  for (int i = 0; i < n; i++) {
    st[i] = node(s[i]);
  }
  st.build();
  for (int i = 0; i < q; i++) {
    int l, r;
    cin >> l >> r;
    cout << st.query(--l, --r) << '\n';
  }
  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;
struct node {
  int state[5][5];
  node() {
    for (int i = 0; i < 5; i++)
      for (int j = 0; j < 5; j++) {
        state[i][j] = 1031231234;
      }
  }
  node(char v) {
    for (int i = 0; i < 5; i++)
      for (int j = 0; j < 5; j++) {
        state[i][j] = (i == j) ? 0 : 1031231234;
      }
    if (v == '2') {
      state[0][0] = 1;
      state[0][1] = 0;
    } else if (v == '0') {
      state[1][1] = 1;
      state[1][2] = 0;
    } else if (v == '1') {
      state[2][2] = 1;
      state[2][3] = 0;
    } else if (v == '6') {
      state[3][3] = 1;
      state[4][4] = 1;
    } else if (v == '7') {
      state[3][3] = 1;
      state[3][4] = 0;
    }
  }
  node combine(const node& o) const {
    node res;
    for (int i = 0; i < 5; i++) {
      for (int j = 0; j < 5; j++) {
        for (int k = 0; k < 5; k++) {
          res.state[i][j] = min(res.state[i][j], state[i][k] + o.state[k][j]);
        }
      }
    }
    return res;
  }
};
struct segmentTree {
  int sz;
  vector<node> elements;
  segmentTree(int s) {
    sz = s;
    elements.resize(2 * sz);
  }
  node& operator[](int i) { return elements[i + sz]; }
  void build() {
    for (int i = sz - 1; i >= 1; i--) {
      elements[i] = elements[i << 1].combine(elements[i << 1 | 1]);
    }
  }
  int query(int l, int r) {
    node lacc, racc;
    lacc.state[0][0] = 0, racc.state[4][4] = 0;
    l += sz;
    r += sz;
    while (l <= r) {
      if (l & 1) {
        lacc = lacc.combine(elements[l]);
        l++;
      }
      if (!(r & 1)) {
        racc = elements[r].combine(racc);
        r--;
      }
      l >>= 1;
      r >>= 1;
    }
    node res = lacc.combine(racc);
    return res.state[0][4] != 1031231234 ? res.state[0][4] : -1;
  }
};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n, q;
  cin >> n >> q;
  string s;
  cin >> s;
  segmentTree st(n);
  for (int i = 0; i < n; i++) {
    st[i] = node(s[i]);
  }
  st.build();
  for (int i = 0; i < q; i++) {
    int l, r;
    cin >> l >> r;
    cout << st.query(--l, --r) << '\n';
  }
  return 0;
}

```