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;
const int maxn = 3e5 + 7;
vector<int> G[maxn];
vector<pair<int, int> > seg;
bool vis[maxn];
bool cyc[maxn];
int par[maxn];
void dfs(int x, int pi) {
  vis[x] = true;
  par[x] = pi;
  for (int y : G[x]) {
    if (!vis[y])
      dfs(y, x);
    else {
      if (!cyc[y] && pi != y) {
        int mx = x, mn = x, cur = x;
        cyc[x] = true;
        while (cur != y) {
          cur = par[cur];
          cyc[cur] = true;
          mx = max(mx, cur);
          mn = min(mn, cur);
        }
        seg.push_back(make_pair(mn, mx));
      }
    }
  }
}
set<int> S;
int z[maxn];
long long pref[maxn];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n, m;
  cin >> n >> m;
  fill(vis, vis + maxn, false);
  fill(cyc, cyc + maxn, false);
  fill(par, par + maxn, -1);
  seg.clear();
  for (int i = 0; i < m; i++) {
    int x, y;
    cin >> x >> y;
    G[x].push_back(y);
    G[y].push_back(x);
  }
  for (int i = 1; i <= n; i++)
    if (!vis[i]) dfs(i, -1);
  sort(seg.begin(), seg.end());
  pref[0] = 0;
  S.clear();
  int idx = seg.size();
  for (int j = n; j >= 0; j--) {
    while (idx > 0 && seg[idx - 1].first >= j) {
      idx--;
      S.insert(seg[idx].second);
    }
    z[j] = (S.size() > 0) ? ((*S.begin()) - 1) : n;
  }
  for (int i = 1; i <= n; i++) pref[i] = pref[i - 1] + z[i];
  int q;
  cin >> q;
  while (q--) {
    int l, r;
    cin >> l >> r;
    int lft = l - 1, rgt = r;
    while (lft < rgt) {
      int mid = (lft + rgt + 1) >> 1;
      if (z[mid] < r)
        lft = mid;
      else
        rgt = mid - 1;
    }
    int pos = lft;
    long long ans = (pref[pos] - pref[l - 1]) + 1LL * r * (r - pos);
    ans -= 1LL * r * (r + 1) / 2 - 1LL * l * (l - 1) / 2;
    ans += r - l + 1;
    cout << ans << "\n";
  }
}
