#include <bits/stdc++.h>
const long long INF = 4e18L + 1;
const int IINF = 1e9 + 1;
const int limit = 1048576;
using namespace std;
template <class TH>
void _dbg(const char *sdbg, TH h) {
  cerr << sdbg << '=' << h << endl;
}
template <class TH, class... TA>
void _dbg(const char *sdbg, TH h, TA... a) {
  while (*sdbg != ',') cerr << *sdbg++;
  cerr << '=' << h << ',';
  _dbg(sdbg + 1, a...);
}
struct Node {
  int dp[5][5];
  Node() {
    for (int i = 0; i < 5; i++) {
      for (int j = 0; j < 5; j++) {
        dp[i][j] = IINF;
      }
    }
  }
};
const int P = 1024 * 256;
Node drzewo[2 * P];
Node merge(Node &a, Node &b) {
  Node ret;
  for (int i = 0; i < 5; i++) {
    for (int j = i; j < 5; j++) {
      for (int k = j; k < 5; k++) {
        ret.dp[i][k] = min(ret.dp[i][k], a.dp[i][j] + b.dp[j][k]);
      }
    }
  }
  return ret;
}
Node getdp(int l, int r) {
  l += P;
  r += P;
  Node retl;
  Node retr;
  for (int i = 0; i < 5; i++) {
    retl.dp[i][i] = 0;
    retr.dp[i][i] = 0;
  }
  while (l + 1 < r) {
    if (l % 2 == 1) {
      retl = merge(retl, drzewo[l]);
      (l);
      l++;
    }
    if (r % 2 == 0) {
      retr = merge(drzewo[r], retr);
      (r);
      r--;
    }
    l /= 2;
    r /= 2;
  }
  (l);
  retl = merge(retl, drzewo[l]);
  if (l != r) {
    (r);
    retr = merge(drzewo[r], retr);
  }
  return merge(retl, retr);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n, q;
  string s;
  cin >> n >> q >> s;
  string t = "2017";
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < 5; j++) {
      drzewo[P + i].dp[j][j] = 0;
    }
    for (int j = 0; j < 4; j++) {
      if (s[i] == t[j]) {
        drzewo[P + i].dp[j][j] = 1;
        drzewo[P + i].dp[j][j + 1] = 0;
      }
    }
    if (s[i] == '6') {
      drzewo[P + i].dp[3][3] = 1;
      drzewo[P + i].dp[4][4] = 1;
    }
  }
  for (int i = P - 1; i > 0; i--) {
    drzewo[i] = merge(drzewo[2 * i], drzewo[2 * i + 1]);
  }
  for (int k = 1; k < 16; k++) {
    (k);
    for (int i = 0; i < 5; i++) {
      for (int j = 0; j < 5; j++) {
        if (0) cout << drzewo[k].dp[i][j] << " ";
      }
      if (0) cout << endl;
    }
    if (0) cout << endl;
  }
  while (q--) {
    int l, r;
    cin >> l >> r;
    l--;
    r--;
    auto w = getdp(l, r);
    if (w.dp[0][4] > 1e6) {
      cout << "-1\n";
    } else {
      cout << w.dp[0][4] << "\n";
    }
  }
}

### Prompt

Your challenge is to write 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>
const long long INF = 4e18L + 1;
const int IINF = 1e9 + 1;
const int limit = 1048576;
using namespace std;
template <class TH>
void _dbg(const char *sdbg, TH h) {
  cerr << sdbg << '=' << h << endl;
}
template <class TH, class... TA>
void _dbg(const char *sdbg, TH h, TA... a) {
  while (*sdbg != ',') cerr << *sdbg++;
  cerr << '=' << h << ',';
  _dbg(sdbg + 1, a...);
}
struct Node {
  int dp[5][5];
  Node() {
    for (int i = 0; i < 5; i++) {
      for (int j = 0; j < 5; j++) {
        dp[i][j] = IINF;
      }
    }
  }
};
const int P = 1024 * 256;
Node drzewo[2 * P];
Node merge(Node &a, Node &b) {
  Node ret;
  for (int i = 0; i < 5; i++) {
    for (int j = i; j < 5; j++) {
      for (int k = j; k < 5; k++) {
        ret.dp[i][k] = min(ret.dp[i][k], a.dp[i][j] + b.dp[j][k]);
      }
    }
  }
  return ret;
}
Node getdp(int l, int r) {
  l += P;
  r += P;
  Node retl;
  Node retr;
  for (int i = 0; i < 5; i++) {
    retl.dp[i][i] = 0;
    retr.dp[i][i] = 0;
  }
  while (l + 1 < r) {
    if (l % 2 == 1) {
      retl = merge(retl, drzewo[l]);
      (l);
      l++;
    }
    if (r % 2 == 0) {
      retr = merge(drzewo[r], retr);
      (r);
      r--;
    }
    l /= 2;
    r /= 2;
  }
  (l);
  retl = merge(retl, drzewo[l]);
  if (l != r) {
    (r);
    retr = merge(drzewo[r], retr);
  }
  return merge(retl, retr);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n, q;
  string s;
  cin >> n >> q >> s;
  string t = "2017";
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < 5; j++) {
      drzewo[P + i].dp[j][j] = 0;
    }
    for (int j = 0; j < 4; j++) {
      if (s[i] == t[j]) {
        drzewo[P + i].dp[j][j] = 1;
        drzewo[P + i].dp[j][j + 1] = 0;
      }
    }
    if (s[i] == '6') {
      drzewo[P + i].dp[3][3] = 1;
      drzewo[P + i].dp[4][4] = 1;
    }
  }
  for (int i = P - 1; i > 0; i--) {
    drzewo[i] = merge(drzewo[2 * i], drzewo[2 * i + 1]);
  }
  for (int k = 1; k < 16; k++) {
    (k);
    for (int i = 0; i < 5; i++) {
      for (int j = 0; j < 5; j++) {
        if (0) cout << drzewo[k].dp[i][j] << " ";
      }
      if (0) cout << endl;
    }
    if (0) cout << endl;
  }
  while (q--) {
    int l, r;
    cin >> l >> r;
    l--;
    r--;
    auto w = getdp(l, r);
    if (w.dp[0][4] > 1e6) {
      cout << "-1\n";
    } else {
      cout << w.dp[0][4] << "\n";
    }
  }
}

```