给你一个长为n的序列a和一个常数k

有m次询问，每次查询一个区间[l,r]内所有数最少分成多少个连续段，使得每段的和都 <= k

如果这一次查询无解，输出"Chtholly"

#define ll long long

const ll maxn = 1e6+5;

const double pi = acos(-1.0);

const ll inf = 0x3f3f3f3f;

ll n, m, k;

ll st[maxn][25];

ll a[maxn];

ll sum[maxn], cnt[maxn];

ll l, r;

ll LOG[maxn];

void init(){

    for(ll i = 1; i <= 20; i++){

        for(ll j = 1; j <= n; j++){

            //st[i][j] = st[st[i][j-1]][j-1];

            st[j][i] = st[st[j][i-1]][i-1];

        }

    }

}

int main() {

    //freopen("in.txt", "r", stdin);

    //freopen("out.txt", "w", stdout);

    cin >>n >> m >> k;

    for(ll i = 1; i <= n; i++){

        scanf("%lld", &a[i]);

        sum[i] = sum[i-1] + a[i];

        cnt[i] = cnt[i-1];

        if (a[i] > k) cnt[i]++;

    }

    for(ll i = 1; i <= n; i++){

        ll pos = upper_bound(sum+1, sum+1+n, sum[i-1]+k)-sum;

        st[i][0] = pos;

       //prllf("%d ", pos);

    }

    init();

    //prllf("---- %d\n", st[2][0]);

    for(ll i = 1; i <= m; i++){

        scanf("%lld%lld", &l, &r);

        if (cnt[r]-cnt[l-1] > 0) printf("Chtholly\n");

        else {

            ll ans = 0, p = l;

            for(ll j = 20; j >= 0; j--){

                if(st[p][j] && st[p][j] <= r){

                    ans += 1<<j;

                    p = st[p][j];

                }

            }

            printf("%lld\n", ans+1);

        }

    }

    return 0;

}