#include <bits/stdc++.h>

#define M_PI 3.14159265358979323846

using namespace std;

//typedef
//-------------------------#include <bits/stdc++.h>

#define M_PI 3.14159265358979323846

using namespace std;

//conversion
//------------------------------------------
inline int toInt(string s)
{
    int v;
    istringstream sin(s);
    sin >> v;
    return v;
}
template <class T>
inline string toString(T x)
{
    ostringstream sout;
    sout << x;
    return sout.str();
}
inline int readInt()
{
    int x;
    scanf("%d", &x);
    return x;
}

//typedef
//------------------------------------------
typedef vector<int> VI;
typedef vector<VI> VVI;
typedef vector<string> VS;
typedef pair<int, int> PII;
typedef pair<int, PII> TIII;
typedef long long LL;
typedef unsigned long long ULL;
typedef vector<LL> VLL;
typedef vector<VLL> VVLL;

//container util

//------------------------------------------
#define ALL(a) (a).begin(), (a).end()
#define RALL(a) (a).rbegin(), (a).rend()
#define PB push_back
#define MP make_pair
#define SZ(a) int((a).size())
#define SQ(a) ((a) * (a))
#define EACH(i, c) for (typeof((c).begin()) i = (c).begin(); i != (c).end(); ++i)
#define EXIST(s, e) ((s).find(e) != (s).end())
#define SORT(c) sort((c).begin(), (c).end())

//repetition
//------------------------------------------
#define FOR(i, s, n) for (int i = s; i < (int)n; ++i)
#define REP(i, n) FOR(i, 0, n)
//#define MOD 1000000007

#define rep(i, a, b) for (int i = a; i < (b); ++i)
#define trav(a, x) for (auto &a : x)
#define all(x) x.begin(), x.end()
#define sz(x) (int)(x).size()

typedef long long ll;
typedef pair<int, int> pii;
typedef vector<int> vi;
const double EPS = 1E-8;

#define chmin(x, y) x = min(x, y)
#define chmax(x, y) x = max(x, y)
const int INF = 2147483647;

const int MAX = 510000;
const int MOD = 1000000007;

long long fac[MAX], finv[MAX], inv[MAX];

// テーブルを作る前処理
void COMinit()
{
    fac[0] = fac[1] = 1;
    finv[0] = finv[1] = 1;
    inv[1] = 1;
    for (int i = 2; i < MAX; i++)
    {
        fac[i] = fac[i - 1] * i % MOD;
        inv[i] = MOD - inv[MOD % i] * (MOD / i) % MOD;
        finv[i] = finv[i - 1] * inv[i] % MOD;
    }
}

// 二項係数計算
long long COM(int n, int k)
{
    if (n < k)
        return 0;
    if (n < 0 || k < 0)
        return 0;
    return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
}

ll S, T, D;
vector<ll> w(100010);
ll sum = 0;
ll mi = (1LL << 60);
bool C(ll x)
{
    ll ts = S + mi;
    ts += sum * (x / D);

    if (ts <= T)
        return true;
    else
        return false;
}
int main()
{
    cin.tie(0);
    ios::sync_with_stdio(false);
    cout << fixed << setprecision(15);

    cin >> S >> T >> D;
    for (int i = 0; i < D; i++)
    {
        cin >> w[i];
        sum += w[i];
        if (S + sum <= T)
        {
            cout << i + 1 << endl;
            return 0;
        }
        mi = min(mi, sum);
    }

    if (sum >= 0)
    {
        cout << -1 << endl;
        return 0;
    }
    else
    {

        ll lb = 0, ub = 100000000100;
        while (ub - lb > 1)
        {
            ll mid = (lb + ub) / 2;
            if (C(mid))
            {
                ub = mid;
            }
            else
            {
                lb = mid;
            }
        }
        ll ans = (1LL<<60);
        //cout << ub << endl;
        
        
        ll ts = S + sum * (max(0LL,(ub - D)) / D);
        //cout << ts << endl;
        for (ll i = 0; i < 2*D; i++)
        {
            ts += w[i%D];
            if (ts <= T)
            {
                ans = min(ans, ub - D + i + 1);
            }
        }
        cout << ans << endl;
    }

    return 0;
}


### Prompt

Develop a solution in cpp to the problem described below:

Chokudai loves eating so much. However, his doctor Akensho told him that he was overweight, so he finally decided to lose his weight.

Chokudai made a slimming plan of a $D$-day cycle. It is represented by $D$ integers $w_0, ..., w_{D-1}$. His weight is $S$ on the 0-th day of the plan and he aims to reduce it to $T$ ($S > T$). If his weight on the $i$-th day of the plan is $x$, it will be $x + w_{i\%D}$ on the $(i+1)$-th day. Note that $i\%D$ is the remainder obtained by dividing $i$ by $D$. If his weight successfully gets less than or equal to $T$, he will stop slimming immediately.

If his slimming plan takes too many days or even does not end forever, he should reconsider it.

Determine whether it ends or not, and report how many days it takes if it ends.



Input

The input consists of a single test case formatted as follows.


$S$ $T$ $D$
$w_0 ... w_{D-1}$


The first line consists of three integers $S$, $T$, $D$ ($1 \leq S, T, D \leq 100,000, S > T$). The second line consists of $D$ integers $w_0, ..., w_{D-1}$ ($-100,000 \leq w_i \leq 100,000$ for each $i$).

Output

If Chokudai's slimming plan ends on the $d$-th day, print $d$ in one line. If it never ends, print $-1$.

Examples

Input

65 60 3
-2 3 -4


Output

4


Input

65 60 3
-2 10 -3


Output

-1


Input

100000 1 1
-1


Output

99999


Input

60 59 1
-123


Output

1

### Response

```cpp
#include <bits/stdc++.h>

#define M_PI 3.14159265358979323846

using namespace std;

//typedef
//-------------------------#include <bits/stdc++.h>

#define M_PI 3.14159265358979323846

using namespace std;

//conversion
//------------------------------------------
inline int toInt(string s)
{
    int v;
    istringstream sin(s);
    sin >> v;
    return v;
}
template <class T>
inline string toString(T x)
{
    ostringstream sout;
    sout << x;
    return sout.str();
}
inline int readInt()
{
    int x;
    scanf("%d", &x);
    return x;
}

//typedef
//------------------------------------------
typedef vector<int> VI;
typedef vector<VI> VVI;
typedef vector<string> VS;
typedef pair<int, int> PII;
typedef pair<int, PII> TIII;
typedef long long LL;
typedef unsigned long long ULL;
typedef vector<LL> VLL;
typedef vector<VLL> VVLL;

//container util

//------------------------------------------
#define ALL(a) (a).begin(), (a).end()
#define RALL(a) (a).rbegin(), (a).rend()
#define PB push_back
#define MP make_pair
#define SZ(a) int((a).size())
#define SQ(a) ((a) * (a))
#define EACH(i, c) for (typeof((c).begin()) i = (c).begin(); i != (c).end(); ++i)
#define EXIST(s, e) ((s).find(e) != (s).end())
#define SORT(c) sort((c).begin(), (c).end())

//repetition
//------------------------------------------
#define FOR(i, s, n) for (int i = s; i < (int)n; ++i)
#define REP(i, n) FOR(i, 0, n)
//#define MOD 1000000007

#define rep(i, a, b) for (int i = a; i < (b); ++i)
#define trav(a, x) for (auto &a : x)
#define all(x) x.begin(), x.end()
#define sz(x) (int)(x).size()

typedef long long ll;
typedef pair<int, int> pii;
typedef vector<int> vi;
const double EPS = 1E-8;

#define chmin(x, y) x = min(x, y)
#define chmax(x, y) x = max(x, y)
const int INF = 2147483647;

const int MAX = 510000;
const int MOD = 1000000007;

long long fac[MAX], finv[MAX], inv[MAX];

// テーブルを作る前処理
void COMinit()
{
    fac[0] = fac[1] = 1;
    finv[0] = finv[1] = 1;
    inv[1] = 1;
    for (int i = 2; i < MAX; i++)
    {
        fac[i] = fac[i - 1] * i % MOD;
        inv[i] = MOD - inv[MOD % i] * (MOD / i) % MOD;
        finv[i] = finv[i - 1] * inv[i] % MOD;
    }
}

// 二項係数計算
long long COM(int n, int k)
{
    if (n < k)
        return 0;
    if (n < 0 || k < 0)
        return 0;
    return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
}

ll S, T, D;
vector<ll> w(100010);
ll sum = 0;
ll mi = (1LL << 60);
bool C(ll x)
{
    ll ts = S + mi;
    ts += sum * (x / D);

    if (ts <= T)
        return true;
    else
        return false;
}
int main()
{
    cin.tie(0);
    ios::sync_with_stdio(false);
    cout << fixed << setprecision(15);

    cin >> S >> T >> D;
    for (int i = 0; i < D; i++)
    {
        cin >> w[i];
        sum += w[i];
        if (S + sum <= T)
        {
            cout << i + 1 << endl;
            return 0;
        }
        mi = min(mi, sum);
    }

    if (sum >= 0)
    {
        cout << -1 << endl;
        return 0;
    }
    else
    {

        ll lb = 0, ub = 100000000100;
        while (ub - lb > 1)
        {
            ll mid = (lb + ub) / 2;
            if (C(mid))
            {
                ub = mid;
            }
            else
            {
                lb = mid;
            }
        }
        ll ans = (1LL<<60);
        //cout << ub << endl;
        
        
        ll ts = S + sum * (max(0LL,(ub - D)) / D);
        //cout << ts << endl;
        for (ll i = 0; i < 2*D; i++)
        {
            ts += w[i%D];
            if (ts <= T)
            {
                ans = min(ans, ub - D + i + 1);
            }
        }
        cout << ans << endl;
    }

    return 0;
}


```