You are given an undirected graph with n vertices. There are no edge-simple cycles with the even length in it. In other words, there are no cycles of even length that pass each edge at most once. Let's enumerate vertices from 1 to n. 

You have to answer q queries. Each query is described by a segment of vertices [l; r], and you have to count the number of its subsegments [x; y] (l ≤ x ≤ y ≤ r), such that if we delete all vertices except the segment of vertices [x; y] (including x and y) and edges between them, the resulting graph is bipartite.

Input

The first line contains two integers n and m (1 ≤ n ≤ 3·105, 1 ≤ m ≤ 3·105) — the number of vertices and the number of edges in the graph.

The next m lines describe edges in the graph. The i-th of these lines contains two integers ai and bi (1 ≤ ai, bi ≤ n; ai ≠ bi), denoting an edge between vertices ai and bi. It is guaranteed that this graph does not contain edge-simple cycles of even length.

The next line contains a single integer q (1 ≤ q ≤ 3·105) — the number of queries.

The next q lines contain queries. The i-th of these lines contains two integers li and ri (1 ≤ li ≤ ri ≤ n) — the query parameters.

Output

Print q numbers, each in new line: the i-th of them should be the number of subsegments [x; y] (li ≤ x ≤ y ≤ ri), such that the graph that only includes vertices from segment [x; y] and edges between them is bipartite.

Examples

Input

6 6
1 2
2 3
3 1
4 5
5 6
6 4
3
1 3
4 6
1 6


Output

5
5
14


Input

8 9
1 2
2 3
3 1
4 5
5 6
6 7
7 8
8 4
7 2
3
1 8
1 4
3 8


Output

27
8
19

Note

The first example is shown on the picture below:

<image>

For the first query, all subsegments of [1; 3], except this segment itself, are suitable.

For the first query, all subsegments of [4; 6], except this segment itself, are suitable.

For the third query, all subsegments of [1; 6] are suitable, except [1; 3], [1; 4], [1; 5], [1; 6], [2; 6], [3; 6], [4; 6].

The second example is shown on the picture below:

<image>
#include <bits/stdc++.h>
using namespace std;
vector<int> pa, dfn, rt, uncut;
vector<vector<int> > g;
int n, m, q, dfntot;
vector<long long> pre;
vector<pair<int, int> > s;
void dfs(int u) {
  dfn[u] = ++dfntot;
  for (auto v : g[u]) {
    if (v == pa[u]) continue;
    if (!dfn[v]) {
      pa[v] = u;
      dfs(v);
    } else if (dfn[v] < dfn[u]) {
      int mn = u, mx = u, x = u;
      while (x != v) {
        x = pa[x];
        mn = min(mn, x);
        mx = max(mx, x);
      }
      rt[mn] = mx;
    }
  }
}
int main() {
  cout.sync_with_stdio(false);
  cin.sync_with_stdio(false);
  cout.tie(0);
  cin.tie(0);
  cin >> n >> m;
  g.resize(n + 1);
  pa.resize(n + 1, 0);
  pre.resize(n + 1, 0);
  dfn.resize(n + 1, 0);
  uncut.resize(n + 1, 0);
  rt.resize(n + 1, n + 1);
  for (int i = 1; i <= m; ++i) {
    int u, v;
    cin >> u >> v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  for (int i = 1; i <= n; ++i)
    if (!dfn[i]) dfs(i);
  int mn = n;
  for (int i = n; i >= 1; --i) {
    mn = min(mn, rt[i]);
    if (rt[i] <= mn) s.push_back(pair<int, int>(i, rt[i]));
  }
  reverse(s.begin(), s.end());
  s.push_back(pair<int, int>(n + 1, n + 1));
  int p = 0, ban = s[0].second;
  for (int i = 1; i <= n; ++i) {
    if (i > s[p].first) ban = s[++p].second;
    pre[i] = pre[i - 1] + ban - i;
    uncut[i] = uncut[i - 1] + (i == s[uncut[i - 1]].second);
  }
  cin >> q;
  while (q--) {
    int l, r;
    cin >> l >> r;
    int k = uncut[r];
    if (k == 0 || s[k - 1].first < l)
      cout << (long long)(r - l + 1) * (r - l + 2) / 2 << endl;
    else
      cout << pre[s[k - 1].first] - pre[l - 1] +
                  (long long)(r - s[k - 1].first) * (r - s[k - 1].first + 1) / 2
           << endl;
  }
  return 0;
}
