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;
void openFile() {
  ios_base ::sync_with_stdio(false);
  cin.tie(NULL);
}
const int maxn = 3e5 + 5;
const int maxm = 1e6 + 5;
const long long inf = 1e9 + 7;
int N, M;
vector<int> vertex[maxn];
int vis[maxn], low[maxn], num[maxn];
int nCycle;
int cntDFS;
vector<int> cycle[maxn];
int inCycle[maxn];
int lef[maxn];
long long sl[maxn];
long long s[maxn];
void enter() {
  cin >> N >> M;
  for (int i = 1; i <= M; ++i) {
    int u, v;
    cin >> u >> v;
    vertex[u].push_back(v);
    vertex[v].push_back(u);
  }
}
void dfs(int u, int p) {
  for (int v : vertex[u]) {
    if (v == p) continue;
    if (!vis[v]) {
      vis[v] = u;
      dfs(v, u);
    } else if (!inCycle[v]) {
      ++nCycle;
      inCycle[v] = nCycle;
      cycle[nCycle].push_back(v);
      int uu = u;
      while (uu != v) {
        inCycle[uu] = nCycle;
        cycle[nCycle].push_back(uu);
        uu = vis[uu];
      }
    }
  }
}
void prepare() {
  lef[1] = 1;
  sl[1] = 1;
  s[1] = 1;
  for (int i = 2; i <= N; ++i) {
    int nc = inCycle[i];
    if (nc == 0)
      lef[i] = lef[i - 1];
    else {
      if (i == *cycle[nc].rbegin()) {
        lef[i] = max(lef[i - 1], *cycle[nc].begin() + 1);
      } else {
        lef[i] = lef[i - 1];
      }
    }
    s[i] = s[i - 1] + i;
    sl[i] = sl[i - 1] + i - lef[i] + 1;
  }
}
long long sum(int l, int r) { return s[r] - s[l - 1]; }
long long sumlef(int l, int r) { return sl[r] - sl[l - 1]; }
void solve() {
  for (int u = 1; u <= N; ++u) {
    if (vis[u] == 0) {
      vis[u] = -1;
      dfs(u, -1);
    }
  }
  for (int i = 1; i <= nCycle; ++i) {
    sort(cycle[i].begin(), cycle[i].end());
  }
  prepare();
  int Q;
  cin >> Q;
  for (int i = 1; i <= Q; ++i) {
    int u, v;
    cin >> u >> v;
    int lo = u, hi = v, mid;
    while (lo <= hi) {
      mid = (lo + hi) / 2;
      if (lef[mid] < u) {
        lo = mid + 1;
      } else {
        hi = mid - 1;
      }
    }
    int last = hi;
    long long res = 0;
    if (last >= u) {
      res = res + sum(1, last - u + 1);
    }
    if (last + 1 <= v) {
      res = res + sumlef(last + 1, v);
    }
    cout << res << endl;
  }
}
int main() {
  openFile();
  enter();
  solve();
  return 0;
}
