为什么60pts

本地能过大样例，跟题解对拍很多组也没有问题

#include<bits/stdc++.h>
#define int long long
#define maxn 200005
#define multicase() int t;cin>>t;while(t--)
using namespace std;
const int mod = 1e9 + 7;
int n,m,v;
map<int,int> a;
vector<int> b;
int ksm(int x,int y)
{
    int res = 1;
    while(y)
    {
        if(y&1) (res *= x) %= mod;
        y>>= 1;
        (x *= x) %= mod;
    }
    return res;
}
signed main()
{
    ios::sync_with_stdio(0);
    cin.tie(0);cout.tie(0);
    multicase()
    {
        a.clear();
        b.clear();
        cin >> n >> m >> v;
        bool flag = 0;
        for(int i=1;i<=m;i++){
            int x,y;
            cin >> x >> y;
            if(a[x]) 
            {
                if(a[x] != y) flag = 1;
            }
            else a[x] = y,b.push_back(x);
        }
        if(flag)
        {
            cout << 0 << "\n";
            continue;
        }
        sort(b.begin(),b.end());
        bool noval = 1;
        int lst = 1,ans = 1;
        if(b.empty())
        {
            ans = ksm(v,2*n-2);
            ans %= mod;
            cout << ans << "\n";
            continue;
        }
        ans = ksm(v,2*(b[0]-1)) * ksm(v,2*(n-b[b.size()-1]));
        ans %= mod;
        for(int i=1;i<b.size();i++)
        {
//          cout << "i = " << i << "\n";
            ans *= ((ksm(v,2*(b[i]-b[i-1])) - ksm(v,(b[i]-b[i-1])) + ksm(v,(b[i]-b[i-1]-1))))%mod;
            ans %= mod;
        }
        ans %= mod;
        cout << ans << "\n";
    }
    return 0;
}