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 N = 300000;
const int inf = (int)1e9 + 1;
const long long big = (long long)1e18 + 1;
const int P = 239;
const int MOD = (int)1e9 + 7;
const int MOD1 = (int)1e9 + 9;
const long double eps = 1e-9;
const long double pi = atan2(0, -1);
const int ABC = 26;
vector<int> g[N];
bool used[N];
int ttt = 0;
int in[N], up[N];
vector<int> cur;
vector<pair<int, int> > restr;
void dfs(int u, int par) {
  used[u] = true;
  in[u] = up[u] = ttt++;
  cur.push_back(u);
  for (int v : g[u])
    if (!used[v]) {
      dfs(v, u);
      up[u] = min(up[u], up[v]);
      if (up[v] == in[v]) {
        int mi = inf, ma = -inf;
        while (true) {
          int w = cur.back();
          cur.pop_back();
          mi = min(mi, w);
          ma = max(ma, w);
          if (w == v) break;
        }
        if (mi != ma) restr.push_back({mi, ma});
      }
    } else if (v != par) {
      up[u] = min(up[u], in[v]);
    }
  if (par == -1) {
    int mi = inf, ma = -inf;
    while ((int)cur.size() > 0) {
      int w = cur.back();
      cur.pop_back();
      mi = min(mi, w);
      ma = max(ma, w);
    }
    if (mi != ma) restr.push_back({mi, ma});
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, m;
  cin >> n >> m;
  for (int i = 0; i < m; i++) {
    int u, v;
    cin >> u >> v;
    u--;
    v--;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  for (int i = 0; i < n; i++)
    if (!used[i]) dfs(i, -1);
  sort(restr.begin(), restr.end(),
       [](pair<int, int> a, pair<int, int> b) { return a.second < b.second; });
  int pos = 0;
  int start[n];
  int cur_max = -1;
  for (int i = 0; i < n; i++) {
    while (pos < (int)restr.size() && restr[pos].second == i)
      cur_max = max(cur_max, restr[pos++].first);
    start[i] = cur_max + 1;
  }
  long long pref[n + 1];
  pref[0] = 0;
  for (int i = 0; i < n; i++) pref[i + 1] = pref[i] + (i - start[i] + 1);
  int q;
  cin >> q;
  while (q--) {
    int x, y;
    cin >> x >> y;
    x--;
    y--;
    long long ans = 0;
    int pos = min(y + 1, (int)(lower_bound(start, start + n, x) - start));
    if (pos - 1 >= x) {
      long long len = (pos - x);
      ans += len * (len + 1) / 2;
    }
    ans += pref[y + 1] - pref[pos];
    cout << ans << "\n";
  }
  return 0;
}
