Problem: You are given n intervals in form [l; r] on a number line.

You are also given m queries in form [x; y]. What is the minimal number of intervals you have to take so that every point (not necessarily integer) from x to y is covered by at least one of them? 

If you can't choose intervals so that every point from x to y is covered, then print -1 for that query.

Input

The first line contains two integers n and m (1 ≤ n, m ≤ 2 ⋅ 10^5) — the number of intervals and the number of queries, respectively.

Each of the next n lines contains two integer numbers l_i and r_i (0 ≤ l_i < r_i ≤ 5 ⋅ 10^5) — the given intervals.

Each of the next m lines contains two integer numbers x_i and y_i (0 ≤ x_i < y_i ≤ 5 ⋅ 10^5) — the queries.

Output

Print m integer numbers. The i-th number should be the answer to the i-th query: either the minimal number of intervals you have to take so that every point (not necessarily integer) from x_i to y_i is covered by at least one of them or -1 if you can't choose intervals so that every point from x_i to y_i is covered.

Examples

Input


2 3
1 3
2 4
1 3
1 4
3 4


Output


1
2
1


Input


3 4
1 3
1 3
4 5
1 2
1 3
1 4
1 5


Output


1
1
-1
-1

Note

In the first example there are three queries:

  1. query [1; 3] can be covered by interval [1; 3]; 
  2. query [1; 4] can be covered by intervals [1; 3] and [2; 4]. There is no way to cover [1; 4] by a single interval; 
  3. query [3; 4] can be covered by interval [2; 4]. It doesn't matter that the other points are covered besides the given query. 



In the second example there are four queries:

  1. query [1; 2] can be covered by interval [1; 3]. Note that you can choose any of the two given intervals [1; 3]; 
  2. query [1; 3] can be covered by interval [1; 3]; 
  3. query [1; 4] can't be covered by any set of intervals; 
  4. query [1; 5] can't be covered by any set of intervals. Note that intervals [1; 3] and [4; 5] together don't cover [1; 5] because even non-integer points should be covered. Here 3.5, for example, isn't covered. 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
int MOD = 1e9 + 7;
const int N = 600000;
int ss[N];
int sd[N][20];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, m;
  cin >> n >> m;
  for (int i = 0; i < n; i++) {
    int l, r;
    cin >> l >> r;
    ss[l] = max(ss[l], r);
  }
  for (int i = 1; i < N; i++) {
    ss[i] = max(ss[i - 1], ss[i]);
  }
  for (int i = 0; i < 20; i++) {
    for (int j = 0; j < N; j++) {
      if (!i) {
        sd[j][i] = ss[j];
      } else {
        sd[j][i] = sd[sd[j][i - 1]][i - 1];
      }
    }
  }
  for (int i = 0; i < m; i++) {
    int x, y;
    cin >> x >> y;
    int nw = x;
    int kol = 0;
    int gv = 0;
    while (nw < y) {
      if (ss[nw] >= y) {
        nw = ss[nw];
        kol++;
        continue;
      }
      int L = 0, R = 20;
      while (R - L > 1) {
        int mid = (R + L) / 2;
        if (sd[nw][mid] >= y) {
          R = mid;
        } else {
          L = mid;
        }
      }
      if (nw >= sd[nw][L]) {
        cout << -1 << endl;
        gv = 1;
        break;
      }
      nw = sd[nw][L];
      kol += (1 << L);
    }
    if (!gv) cout << kol << endl;
  }
}