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>
using namespace std;
template <class T1, class T2>
inline void chkmin(T1 &x, const T2 &y) {
  if (x > y) x = y;
}
template <class T1, class T2>
inline void chkmax(T1 &x, const T2 &y) {
  if (x < y) x = y;
}
const int MAXN = 500228;
int n, m;
int l[MAXN], r[MAXN];
int x[MAXN], y[MAXN];
int jump[MAXN][20];
vector<int> st[MAXN];
vector<int> en[MAXN];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m;
  for (int i = 0; i < n; i++) {
    cin >> l[i] >> r[i];
    st[l[i]].push_back(r[i]);
    en[r[i]].push_back(r[i]);
  }
  for (int i = 0; i < m; i++) {
    cin >> x[i] >> y[i];
  }
  multiset<int> cur;
  for (int i = 0; i < MAXN; i++) {
    for (auto x : st[i]) {
      cur.insert(x);
    }
    if (!cur.empty()) {
      jump[i][0] = *cur.rbegin();
    } else {
      jump[i][0] = i;
    }
    for (auto x : en[i]) {
      cur.erase(cur.find(x));
    }
  }
  for (int k = 1; k <= 19; k++) {
    for (int i = 0; i < MAXN; i++) {
      jump[i][k] = jump[jump[i][k - 1]][k - 1];
    }
  }
  for (int i = 0; i < m; i++) {
    int cur = x[i];
    int ans = 0;
    for (int k = 19; k >= 0; k--) {
      if (jump[cur][k] < y[i]) {
        cur = jump[cur][k];
        ans += 1 << k;
      }
    }
    if (jump[cur][0] >= y[i]) {
      ans++;
      cout << ans << '\n';
    } else {
      cout << -1 << '\n';
    }
  }
}