#include <bits/stdc++.h>
using namespace std;
template <class T, class T2>
inline int chkmax(T &x, const T2 &y) {
  return x < y ? x = y, 1 : 0;
}
template <class T, class T2>
inline int chkmin(T &x, const T2 &y) {
  return x > y ? x = y, 1 : 0;
}
const int MAXN = (int)200042;
const int inf = (int)1e9 + 42;
string s;
struct node {
  int dp[5][5];
  node() {
    for (int l = 0; l <= 4; l++)
      for (int r = 0; r <= 4; r++) dp[l][r] = (l == r) ? 0 : inf;
  }
  node(char c) {
    for (int l = 0; l <= 4; l++)
      for (int r = 0; r <= 4; r++) dp[l][r] = (l == r) ? 0 : inf;
    if (c == '2') dp[0][1] = 0, dp[0][0] = 1;
    if (c == '0') dp[1][2] = 0, dp[1][1] = 1;
    if (c == '1') dp[2][3] = 0, dp[2][2] = 1;
    if (c == '7') dp[3][4] = 0, dp[3][3] = 1;
    if (c == '6') dp[3][3] = 1, dp[4][4] = 1;
  }
};
node temp, broken;
node merge(node l, node r) {
  temp = node();
  for (int i = 0; i <= 4; i++) temp.dp[i][i] = inf;
  for (int i = 0; i <= 4; i++)
    for (int k = i; k <= 4; k++)
      for (int j = k; j <= 4; j++)
        chkmin(temp.dp[i][j], l.dp[i][k] + r.dp[k][j]);
  return temp;
}
struct segment_tree {
  node tr[4 * MAXN];
  void init(int l, int r, int idx) {
    if (l == r) {
      tr[idx] = node(s[l]);
      return;
    }
    int mid = (l + r) >> 1;
    init(l, mid, 2 * idx + 1);
    init(mid + 1, r, 2 * idx + 2);
    tr[idx] = merge(tr[2 * idx + 1], tr[2 * idx + 2]);
  }
  node query(int qL, int qR, int l, int r, int idx) {
    if (l > qR || r < qL) return broken;
    if (qL <= l && r <= qR) return tr[idx];
    int mid = (l + r) >> 1;
    return merge(query(qL, qR, l, mid, 2 * idx + 1),
                 query(qL, qR, mid + 1, r, 2 * idx + 2));
  }
};
int n, q;
void read() { cin >> n >> q >> s; }
segment_tree t;
void solve() {
  t.init(0, n - 1, 0);
  while (q--) {
    int l, r;
    cin >> l >> r;
    l--, r--;
    int ans = t.query(l, r, 0, n - 1, 0).dp[0][4];
    if (ans == inf) ans = -1;
    cout << ans << '\n';
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  read();
  solve();
  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;
template <class T, class T2>
inline int chkmax(T &x, const T2 &y) {
  return x < y ? x = y, 1 : 0;
}
template <class T, class T2>
inline int chkmin(T &x, const T2 &y) {
  return x > y ? x = y, 1 : 0;
}
const int MAXN = (int)200042;
const int inf = (int)1e9 + 42;
string s;
struct node {
  int dp[5][5];
  node() {
    for (int l = 0; l <= 4; l++)
      for (int r = 0; r <= 4; r++) dp[l][r] = (l == r) ? 0 : inf;
  }
  node(char c) {
    for (int l = 0; l <= 4; l++)
      for (int r = 0; r <= 4; r++) dp[l][r] = (l == r) ? 0 : inf;
    if (c == '2') dp[0][1] = 0, dp[0][0] = 1;
    if (c == '0') dp[1][2] = 0, dp[1][1] = 1;
    if (c == '1') dp[2][3] = 0, dp[2][2] = 1;
    if (c == '7') dp[3][4] = 0, dp[3][3] = 1;
    if (c == '6') dp[3][3] = 1, dp[4][4] = 1;
  }
};
node temp, broken;
node merge(node l, node r) {
  temp = node();
  for (int i = 0; i <= 4; i++) temp.dp[i][i] = inf;
  for (int i = 0; i <= 4; i++)
    for (int k = i; k <= 4; k++)
      for (int j = k; j <= 4; j++)
        chkmin(temp.dp[i][j], l.dp[i][k] + r.dp[k][j]);
  return temp;
}
struct segment_tree {
  node tr[4 * MAXN];
  void init(int l, int r, int idx) {
    if (l == r) {
      tr[idx] = node(s[l]);
      return;
    }
    int mid = (l + r) >> 1;
    init(l, mid, 2 * idx + 1);
    init(mid + 1, r, 2 * idx + 2);
    tr[idx] = merge(tr[2 * idx + 1], tr[2 * idx + 2]);
  }
  node query(int qL, int qR, int l, int r, int idx) {
    if (l > qR || r < qL) return broken;
    if (qL <= l && r <= qR) return tr[idx];
    int mid = (l + r) >> 1;
    return merge(query(qL, qR, l, mid, 2 * idx + 1),
                 query(qL, qR, mid + 1, r, 2 * idx + 2));
  }
};
int n, q;
void read() { cin >> n >> q >> s; }
segment_tree t;
void solve() {
  t.init(0, n - 1, 0);
  while (q--) {
    int l, r;
    cin >> l >> r;
    l--, r--;
    int ans = t.query(l, r, 0, n - 1, 0).dp[0][4];
    if (ans == inf) ans = -1;
    cout << ans << '\n';
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  read();
  solve();
  return 0;
}

```