思路：我们由题意可以知道我们只需要维护区间gcd即可，因为差分一下后，维护的差分数组的区间gcd即为原数组所要求的值

#include<bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef pair<ll, ll>PII;
const int N = 2e5 + 10;
const int MOD = 998244353;
const int INF = 0X3F3F3F3F;
const int dx[] = {-1, 1, 0, 0, -1, -1, +1, +1};
const int dy[] = {0, 0, -1, 1, -1, +1, -1, +1};
const int M = 1e9 + 7;

//线段树维护区间gcd模板
ll a[N];
struct node
{
	int l, r;
	int gcd;
}tr[N << 2];

void push(int u)
{
	tr[u].gcd = gcd(tr[u << 1].gcd, tr[u << 1 | 1].gcd);
}
void build(int u, int l, int r)
{
	if(l == r) tr[u] = {l, r, 0};
	else {
		tr[u] = {l, r};
		int mid = tr[u].l + tr[u].r >> 1;
		build(u << 1, l, mid);
		build(u << 1 | 1, mid + 1, r);
		push(u);
	}
}

void modify(int u, int x, int d)
{
	if(tr[u].l == x && tr[u].r == x) tr[u].gcd = d;
	else{
		int mid = tr[u].l + tr[u].r >> 1;
		if(x <= mid) modify(u << 1, x, d);
		else modify(u << 1 | 1, x, d);//单点修改
		push(u);
	}
}

int query(int u, int l, int r)
{
	if(tr[u].l >= l && tr[u].r <= r) return tr[u].gcd;
	else {
		int res = 0;
		int mid = tr[u].l + tr[u].r >> 1;
		if(l <= mid) res = gcd(res, query(u << 1, l, r));
		if(r > mid) res = gcd(res, query(u << 1 | 1, l, r));
		return res;
	}
}
int main()
{
	int t;
	cin >> t;
	while(t --){
	    int n, m;
		cin >> n >> m;
		for(int i = 1; i <= n; i ++) cin >> a[i];
		build(1, 1, n);
		for(int i = 2; i <= n; i ++) modify(1, i, abs(a[i] - a[i - 1]));//维护差分的gcd即可
		while(m --){
			int l, r;
			cin >> l >> r;
			if(l == r) cout << 0 << endl;
			else 
			cout << query(1, l + 1, r) << endl;//注意区间gcd的范围，比如（2，4）应为gcd（3， 4）
		}
		cout << endl;
	}  
}