#include <bits/stdc++.h>
using namespace std;

const int N = 4e5 + 10, M = N << 2; //因为需要装无向图，所以N=2e5,又因为有新图有旧图，都保存到一个数组中，所以需要再乘以2

//链式前向星
int e[M], h1[N], h2[N], idx, w[M], ne[M];
void add(int h[], int a, int b, int c = 0) {
    e[idx] = b, ne[idx] = h[a], w[idx] = c, h[a] = idx++;
}

int f[N][16];
int depth[N], dist[N];

void bfs() {
    // 1号点是源点
    depth[1] = 1;
    queue<int> q;
    q.push(1);
    while (q.size()) {
        int u = q.front();
        q.pop();
        for (int i = h2[u]; ~i; i = ne[i]) {
            int j = e[i];
            if (!depth[j]) {
                q.push(j);
                depth[j] = depth[u] + 1;
                dist[j] = dist[u] + w[i];
                f[j][0] = u;
                for (int k = 1; k <= 15; k++) f[j][k] = f[f[j][k - 1]][k - 1];
            }
        }
    }
}

//最近公共祖先
int lca(int a, int b) {
    if (depth[a] < depth[b]) swap(a, b);
    for (int k = 15; k >= 0; k--)
        if (depth[f[a][k]] >= depth[b]) a = f[a][k];
    if (a == b) return a;
    for (int k = 15; k >= 0; k--)
        if (f[a][k] != f[b][k]) a = f[a][k], b = f[b][k];
    return f[a][0];
}

//边双连通分量
int dfn[N], low[N], ts, stk[N], top;
vector<int> dcc[N]; //边双中有哪些原始点
int id[N], dcc_cnt; //原始点x属于哪个边双连通分量，dcc_cnt指边双连通分量个数
int is_bridge[M];   //记录哪些边是割边
void tarjan(int u, int fa) {
    dfn[u] = low[u] = ++ts;
    stk[++top] = u;
    for (int i = h1[u]; ~i; i = ne[i]) {
        int j = e[i];
        if (!dfn[j]) {
            tarjan(j, i);
            low[u] = min(low[u], low[j]);
            if (dfn[u] < low[j]) is_bridge[i] = is_bridge[i ^ 1] = true; //记录割边
        } else if (i != (fa ^ 1))
            low[u] = min(low[u], dfn[j]);
    }
    if (dfn[u] == low[u]) {
        ++dcc_cnt; //边双数量+1
        int x;
        do {
            x = stk[top--];
            id[x] = dcc_cnt;           // 记录点与边双关系
            dcc[dcc_cnt].push_back(x); // 记录边双中有哪些点
        } while (x != u);
    }
}

int n, m, q;

int main() {
    memset(h1, -1, sizeof h1); // h1是原图的表头
    memset(h2, -1, sizeof h2); // h2是新生成的缩完点的图表头

    scanf("%d %d", &n, &m);

    //原图
    for (int i = 1; i <= m; i++) {
        int a, b;
        scanf("%d %d", &a, &b);
        add(h1, a, b), add(h1, b, a);
    }

    //用tarjan来缩点，将边双连通分量缩点
    for (int i = 1; i <= n; i++)
        if (!dfn[i]) tarjan(i, -1);

    //将新图建出来
    for (int u = 1; u <= n; u++)
        for (int i = h1[u]; ~i; i = ne[i]) {
            int j = e[i];
            if (id[u] != id[j])
                add(h2, id[u], id[j]), add(h2, id[j], id[u]);
        }

    //随便找个存在的号作为根节点，预处理出每个点到根节点的距离
    bfs();

    scanf("%d", &q); // q次询问
    for (int i = 1; i <= q; i++) {
        int a, b;
        scanf("%d %d", &a, &b);
        a = id[a], b = id[b];
        printf("%d\n", depth[a] + depth[b] - depth[lca(a, b)] * 2); //这就很显然了
    }

    return 0;
}