#include <bits/stdc++.h>
using namespace std;
const int MAX = 200005;
string str, bad = "2016";
struct node {
  int dp[4][4], ans[4];
} seg[4 * MAX];
node merge(node a, node b) {
  node ans;
  for (int l = 0; l < 4; l++)
    for (int r = l; r < 4; r++) {
      ans.dp[l][r] = 1e9;
      for (int k = l; k <= r; k++)
        ans.dp[l][r] = min(ans.dp[l][r], a.dp[l][k] + b.dp[k][r]);
    }
  for (int i = 0; i < 4; i++) {
    ans.ans[i] = a.ans[i] + b.dp[3][3];
    for (int k = i; k < 4; k++)
      ans.ans[i] = min(ans.ans[i], a.dp[i][k] + b.ans[k]);
  }
  return ans;
}
void build(int v, int s, int e) {
  if (e - s < 2) {
    for (int l = 0; l < 4; l++)
      for (int r = l; r < 4; r++)
        if (l == r)
          seg[v].dp[l][r] = 0;
        else
          seg[v].dp[l][r] = 1e9;
    for (int i = 0; i < 4; i++) {
      seg[v].ans[i] = 1e9;
      if (str[s] == bad[i]) {
        seg[v].dp[i][i] = 1;
        if (i < 3) seg[v].dp[i][i + 1] = 0;
      }
    }
    if (str[s] == '7') seg[v].ans[3] = 0;
    return;
  }
  int mid = (s + e) / 2;
  build(2 * v, s, mid);
  build(2 * v + 1, mid, e);
  seg[v] = merge(seg[2 * v], seg[2 * v + 1]);
}
node get(int l, int r, int v, int s, int e) {
  if (l <= s && e <= r) return seg[v];
  int mid = (s + e) / 2;
  if (r <= mid) return get(l, r, 2 * v, s, mid);
  if (mid <= l) return get(l, r, 2 * v + 1, mid, e);
  return merge(get(l, r, 2 * v, s, mid), get(l, r, 2 * v + 1, mid, e));
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n, q;
  cin >> n >> q >> str;
  build(1, 0, n);
  while (q--) {
    int l, r;
    cin >> l >> r;
    l--;
    node ans = get(l, r, 1, 0, n);
    if (ans.ans[0] > 5e8)
      cout << -1 << "\n";
    else
      cout << ans.ans[0] << "\n";
  }
  return 0;
}

### Prompt

In cpp, your task is to solve 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 MAX = 200005;
string str, bad = "2016";
struct node {
  int dp[4][4], ans[4];
} seg[4 * MAX];
node merge(node a, node b) {
  node ans;
  for (int l = 0; l < 4; l++)
    for (int r = l; r < 4; r++) {
      ans.dp[l][r] = 1e9;
      for (int k = l; k <= r; k++)
        ans.dp[l][r] = min(ans.dp[l][r], a.dp[l][k] + b.dp[k][r]);
    }
  for (int i = 0; i < 4; i++) {
    ans.ans[i] = a.ans[i] + b.dp[3][3];
    for (int k = i; k < 4; k++)
      ans.ans[i] = min(ans.ans[i], a.dp[i][k] + b.ans[k]);
  }
  return ans;
}
void build(int v, int s, int e) {
  if (e - s < 2) {
    for (int l = 0; l < 4; l++)
      for (int r = l; r < 4; r++)
        if (l == r)
          seg[v].dp[l][r] = 0;
        else
          seg[v].dp[l][r] = 1e9;
    for (int i = 0; i < 4; i++) {
      seg[v].ans[i] = 1e9;
      if (str[s] == bad[i]) {
        seg[v].dp[i][i] = 1;
        if (i < 3) seg[v].dp[i][i + 1] = 0;
      }
    }
    if (str[s] == '7') seg[v].ans[3] = 0;
    return;
  }
  int mid = (s + e) / 2;
  build(2 * v, s, mid);
  build(2 * v + 1, mid, e);
  seg[v] = merge(seg[2 * v], seg[2 * v + 1]);
}
node get(int l, int r, int v, int s, int e) {
  if (l <= s && e <= r) return seg[v];
  int mid = (s + e) / 2;
  if (r <= mid) return get(l, r, 2 * v, s, mid);
  if (mid <= l) return get(l, r, 2 * v + 1, mid, e);
  return merge(get(l, r, 2 * v, s, mid), get(l, r, 2 * v + 1, mid, e));
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n, q;
  cin >> n >> q >> str;
  build(1, 0, n);
  while (q--) {
    int l, r;
    cin >> l >> r;
    l--;
    node ans = get(l, r, 1, 0, n);
    if (ans.ans[0] > 5e8)
      cout << -1 << "\n";
    else
      cout << ans.ans[0] << "\n";
  }
  return 0;
}

```