#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 5;
struct node {
  int m[5][5];
  node(char c = '$') {
    for (__typeof(4) i = 0; i <= 4; i++) {
      for (__typeof(4) j = 0; j <= 4; j++) m[i][j] = 1e8 + 8;
      if (c != '$') m[i][i] = 0;
    }
    if (c == '2') {
      m[0][0] = 1;
      m[0][1] = 0;
    } else if (c == '0') {
      m[1][1] = 1;
      m[1][2] = 0;
    } else if (c == '1') {
      m[2][2] = 1;
      m[2][3] = 0;
    } else if (c == '7') {
      m[3][3] = 1;
      m[3][4] = 0;
    } else if (c == '6') {
      m[3][3] = 1;
      m[4][4] = 1;
    }
  }
};
int n, q;
char s[N];
node tree[4 * N];
inline node merge(node &a, node &b) {
  node ans = node();
  for (__typeof(4) i = 0; i <= 4; i++)
    for (__typeof(4) j = 0; j <= 4; j++)
      for (__typeof(4) k = 0; k <= 4; k++)
        ans.m[i][j] = min(ans.m[i][j], a.m[i][k] + b.m[k][j]);
  return ans;
}
void build(int id, int l, int r) {
  if (l == r) {
    tree[id] = node(s[l]);
    return;
  }
  int mid = (l + r) / 2;
  int tmp = id + id;
  build(tmp, l, mid);
  build(tmp + 1, mid + 1, r);
  tree[id] = merge(tree[tmp], tree[tmp + 1]);
}
node query(int x, int y, int id = 1, int l = 1, int r = n) {
  if (x > r || y < l) return node('3');
  if (x <= l && y >= r) return tree[id];
  int mid = (l + r) / 2;
  int tmp = id + id;
  node left = query(x, y, tmp, l, mid);
  node right = query(x, y, tmp + 1, mid + 1, r);
  return merge(left, right);
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int l, r;
  cin >> n >> q;
  cin >> (s + 1);
  build(1, 1, n);
  while (q--) {
    cin >> l >> r;
    int ans = query(l, r).m[0][4];
    cout << (ans < 1e8 + 8 ? ans : -1) << endl;
  }
  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;
struct node {
  int m[5][5];
  node(char c = '$') {
    for (__typeof(4) i = 0; i <= 4; i++) {
      for (__typeof(4) j = 0; j <= 4; j++) m[i][j] = 1e8 + 8;
      if (c != '$') m[i][i] = 0;
    }
    if (c == '2') {
      m[0][0] = 1;
      m[0][1] = 0;
    } else if (c == '0') {
      m[1][1] = 1;
      m[1][2] = 0;
    } else if (c == '1') {
      m[2][2] = 1;
      m[2][3] = 0;
    } else if (c == '7') {
      m[3][3] = 1;
      m[3][4] = 0;
    } else if (c == '6') {
      m[3][3] = 1;
      m[4][4] = 1;
    }
  }
};
int n, q;
char s[N];
node tree[4 * N];
inline node merge(node &a, node &b) {
  node ans = node();
  for (__typeof(4) i = 0; i <= 4; i++)
    for (__typeof(4) j = 0; j <= 4; j++)
      for (__typeof(4) k = 0; k <= 4; k++)
        ans.m[i][j] = min(ans.m[i][j], a.m[i][k] + b.m[k][j]);
  return ans;
}
void build(int id, int l, int r) {
  if (l == r) {
    tree[id] = node(s[l]);
    return;
  }
  int mid = (l + r) / 2;
  int tmp = id + id;
  build(tmp, l, mid);
  build(tmp + 1, mid + 1, r);
  tree[id] = merge(tree[tmp], tree[tmp + 1]);
}
node query(int x, int y, int id = 1, int l = 1, int r = n) {
  if (x > r || y < l) return node('3');
  if (x <= l && y >= r) return tree[id];
  int mid = (l + r) / 2;
  int tmp = id + id;
  node left = query(x, y, tmp, l, mid);
  node right = query(x, y, tmp + 1, mid + 1, r);
  return merge(left, right);
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int l, r;
  cin >> n >> q;
  cin >> (s + 1);
  build(1, 1, n);
  while (q--) {
    cin >> l >> r;
    int ans = query(l, r).m[0][4];
    cout << (ans < 1e8 + 8 ? ans : -1) << endl;
  }
  return 0;
}

```