#include <bits/stdc++.h>
using namespace std;
const int N = (int)2e5 + 123;
const int INF = (int)1e9 + 123;
int n, q;
char a[N];
struct Node {
  int dp[5][5];
  Node() {
    for (int i = 0; i < 5; i++) {
      for (int j = i; j < 5; j++) {
        dp[i][j] = INF;
        if (i == j) {
          dp[i][j] = 0;
        }
      }
    }
  }
  void init(int v) {
    if (v == 2) {
      dp[0][0] = 1;
      dp[0][1] = 0;
    } else if (v == 0) {
      dp[1][1] = 1;
      dp[1][2] = 0;
    } else if (v == 1) {
      dp[2][2] = 1;
      dp[2][3] = 0;
    } else if (v == 7) {
      dp[3][3] = 1;
      dp[3][4] = 0;
    } else if (v == 6) {
      dp[3][3] = 1;
      dp[4][4] = 1;
    }
  }
};
Node operator+(const Node &a, const Node &b) {
  Node ans;
  for (int i = 0; i < 5; i++) {
    for (int j = 0; j < 5; j++) {
      ans.dp[i][j] = INF;
    }
  }
  for (int i = 0; i < 5; i++) {
    for (int j = i; j < 5; j++) {
      for (int k = i; k <= j; k++) {
        ans.dp[i][j] = min(ans.dp[i][j], a.dp[i][k] + b.dp[k][j]);
      }
    }
  }
  return ans;
}
Node neutral;
Node tree[4 * N];
Node get(int i, int tl, int tr, int l, int r) {
  if (tl >= tr || tl >= r || tr <= l) {
    return neutral;
  }
  if (l <= tl && tr <= r) {
    return tree[i];
  }
  int tm = (tl + tr) / 2;
  return get(2 * i + 1, tl, tm, l, r) + get(2 * i + 2, tm, tr, l, r);
}
void build(int i, int tl, int tr) {
  if (tl >= tr) {
    return;
  }
  if (tl == tr - 1) {
    tree[i].init(a[tl] - '0');
    return;
  }
  int tm = (tl + tr) / 2;
  build(2 * i + 1, tl, tm);
  build(2 * i + 2, tm, tr);
  tree[i] = tree[2 * i + 1] + tree[2 * i + 2];
}
void run() {
  cin >> n >> q;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  build(0, 0, n);
  for (int i = 0; i < q; i++) {
    int l, r;
    cin >> l >> r;
    int ans = get(0, 0, n, l - 1, r).dp[0][4];
    if (ans == INF) {
      cout << "-1\n";
    } else {
      cout << ans << "\n";
    }
  }
}
int main() {
  run();
  return 0;
}

### Prompt

Generate 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 = (int)2e5 + 123;
const int INF = (int)1e9 + 123;
int n, q;
char a[N];
struct Node {
  int dp[5][5];
  Node() {
    for (int i = 0; i < 5; i++) {
      for (int j = i; j < 5; j++) {
        dp[i][j] = INF;
        if (i == j) {
          dp[i][j] = 0;
        }
      }
    }
  }
  void init(int v) {
    if (v == 2) {
      dp[0][0] = 1;
      dp[0][1] = 0;
    } else if (v == 0) {
      dp[1][1] = 1;
      dp[1][2] = 0;
    } else if (v == 1) {
      dp[2][2] = 1;
      dp[2][3] = 0;
    } else if (v == 7) {
      dp[3][3] = 1;
      dp[3][4] = 0;
    } else if (v == 6) {
      dp[3][3] = 1;
      dp[4][4] = 1;
    }
  }
};
Node operator+(const Node &a, const Node &b) {
  Node ans;
  for (int i = 0; i < 5; i++) {
    for (int j = 0; j < 5; j++) {
      ans.dp[i][j] = INF;
    }
  }
  for (int i = 0; i < 5; i++) {
    for (int j = i; j < 5; j++) {
      for (int k = i; k <= j; k++) {
        ans.dp[i][j] = min(ans.dp[i][j], a.dp[i][k] + b.dp[k][j]);
      }
    }
  }
  return ans;
}
Node neutral;
Node tree[4 * N];
Node get(int i, int tl, int tr, int l, int r) {
  if (tl >= tr || tl >= r || tr <= l) {
    return neutral;
  }
  if (l <= tl && tr <= r) {
    return tree[i];
  }
  int tm = (tl + tr) / 2;
  return get(2 * i + 1, tl, tm, l, r) + get(2 * i + 2, tm, tr, l, r);
}
void build(int i, int tl, int tr) {
  if (tl >= tr) {
    return;
  }
  if (tl == tr - 1) {
    tree[i].init(a[tl] - '0');
    return;
  }
  int tm = (tl + tr) / 2;
  build(2 * i + 1, tl, tm);
  build(2 * i + 2, tm, tr);
  tree[i] = tree[2 * i + 1] + tree[2 * i + 2];
}
void run() {
  cin >> n >> q;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  build(0, 0, n);
  for (int i = 0; i < q; i++) {
    int l, r;
    cin >> l >> r;
    int ans = get(0, 0, n, l - 1, r).dp[0][4];
    if (ans == INF) {
      cout << "-1\n";
    } else {
      cout << ans << "\n";
    }
  }
}
int main() {
  run();
  return 0;
}

```