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 = (1e6) + 7;
const int inf = (1e9) + 7;
const long long LLinf = (1e18) + 7;
const long double eps = 1e-9;
const long long mod = 1e9 + 7;
stack<int> stos;
vector<int> wek[maxn];
int numer = 1;
int cykl[maxn];
int stan[maxn];
inline void sciagaj(int x) {
  while (stos.size() && stos.top() != x) {
    cykl[stos.top()] = numer;
    stos.pop();
  }
  if (stos.size()) {
    cykl[stos.top()] = numer;
    stos.pop();
  }
  numer++;
}
void dfs(int x, int skad) {
  stos.push(x);
  stan[x] = 1;
  for (auto s : wek[x]) {
    if (s == skad) continue;
    if (stan[s] == 0) {
      dfs(s, x);
    } else {
      if (stan[s] == 1 && cykl[x] == 0) sciagaj(s);
    }
  }
}
struct pyt {
  int l, p, numer;
  long long odp;
  bool operator<(const pyt a) const { return l < a.l; }
};
pyt zap[maxn];
set<int> cycles;
long long res[maxn];
int odp[maxn];
int maxnumer[maxn];
const int pot = 1 << 19;
long long tree[2 * pot + 7];
long long suma[2 * pot + 7];
void dodaj(int l, int p, long long ile, int wierz = 1, int lewo = 0,
           int prawo = pot - 1) {
  if (lewo > p || prawo < l) return;
  if (l <= lewo && prawo <= p) {
    tree[wierz] += ile;
    suma[wierz] += ile * (prawo - lewo + 1);
    return;
  }
  int sr = (lewo + prawo) / 2;
  dodaj(l, p, ile, 2 * wierz, lewo, sr);
  dodaj(l, p, ile, 2 * wierz + 1, sr + 1, prawo);
  suma[wierz] = suma[2 * wierz] + suma[2 * wierz + 1] +
                tree[wierz] * (long long)(prawo - lewo + 1);
  return;
}
long long sum(int l, int p, int wierz = 1, int lewo = 0, int prawo = pot - 1) {
  if (lewo > p || prawo < l) return 0LL;
  if (l <= lewo && prawo <= p) return suma[wierz];
  int sr = (lewo + prawo) / 2;
  int ll = max(l, lewo);
  int pp = min(p, prawo);
  int dl = pp - ll + 1;
  return tree[wierz] * (long long)dl + sum(l, p, 2 * wierz, lewo, sr) +
         sum(l, p, 2 * wierz + 1, sr + 1, prawo);
}
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 a, b;
    cin >> a >> b;
    wek[a].push_back(b);
    wek[b].push_back(a);
  }
  for (int i = 1; i <= n; i++)
    if (stan[i] == 0) dfs(i, 0);
  for (int i = 1; i <= n; i++) {
    maxnumer[cykl[i]] = i;
  }
  for (int i = 1; i < numer; i++)
    if (maxnumer[i] > 0) cycles.insert(maxnumer[i]);
  for (int i = 1; i <= n; i++) {
    if (cycles.size())
      odp[i] = *cycles.begin();
    else
      odp[i] = n + 1;
    dodaj(i, odp[i] - 1, 1LL);
    cycles.erase(maxnumer[cykl[i]]);
  }
  int q;
  cin >> q;
  for (int i = 0; i < q; i++) {
    cin >> zap[i].l >> zap[i].p;
    zap[i].numer = i;
  }
  sort(zap, zap + q);
  int curr = 1;
  for (int i = 0; i < q; i++) {
    while (curr < zap[i].l) {
      dodaj(curr, odp[curr] - 1, -1);
      curr++;
    }
    res[zap[i].numer] = sum(zap[i].l, zap[i].p);
  }
  for (int i = 0; i < q; i++) cout << res[i] << "\n";
  return 0;
}
