#include <bits/stdc++.h>
using namespace std;
template <typename T1, typename T2>
inline T1 max(T1 a, T2 b) {
  return a < b ? b : a;
}
template <typename T1, typename T2>
inline T1 min(T1 a, T2 b) {
  return a < b ? a : b;
}
namespace ae86 {
const int bufl = 1 << 15;
char buf[bufl], *s = buf, *t = buf;
inline int fetch() {
  if (s == t) {
    t = (s = buf) + fread(buf, 1, bufl, stdin);
    if (s == t) return EOF;
  }
  return *s++;
}
inline long long ty() {
  long long a = 0;
  int b = 1, c = fetch();
  while (!isdigit(c)) b ^= c == '-', c = fetch();
  while (isdigit(c)) a = a * 10 + c - 48, c = fetch();
  return b ? a : -a;
}
}  // namespace ae86
using ae86::ty;
const int _ = 1000007, mo = 1000000007;
template <typename T1, typename T2>
inline T1 ad(T1 &a, T2 b) {
  return a = a + b >= mo ? a + b - mo : a + b;
}
template <typename T1, typename T2>
inline T1 dl(T1 &a, T2 b) {
  return a = a >= b ? a - b : a - b + mo;
}
inline long long powa(long long a, long long t) {
  long long b = 1;
  while (t) {
    if (t & 1) b = b * a % mo;
    a = a * a % mo, t >>= 1;
  }
  return b;
}
inline long long inva(long long a) { return powa(a, mo - 2); }
int n, tar, val[_], pval[_], ha[_] = {0}, lha = 0, cnt[_] = {0}, scnt[_] = {0},
                             loca[_] = {0};
long long len, f[_] = {0}, sf[_] = {0};
int main() {
  n = ty(), len = ty(), tar = ty();
  for (int i = 1; i <= n; i++) val[i] = ty(), ha[++lha] = val[i];
  sort(ha + 1, ha + lha + 1), lha = unique(ha + 1, ha + lha + 1) - ha - 1;
  for (int i = 1; i <= n; i++)
    val[i] = pval[i] = lower_bound(ha + 1, ha + lha + 1, val[i]) - ha,
    cnt[val[i]]++;
  for (int i = 1; i <= lha; i++) scnt[i] = scnt[i - 1] + cnt[i];
  sort(pval + 1, pval + n + 1);
  long long ans = len % mo, duc = len / n, dms = len % n;
  for (int i = 1; i <= n; i++) f[i] = 1;
  for (int i = 2; i <= tar; i++) {
    for (int j = 1; j <= n; j++) sf[j] = (sf[j - 1] + f[j]) % mo;
    long long dlt = 0;
    for (int j = 1; j <= n; j++) f[j] = sf[scnt[pval[j]]], ad(dlt, f[j]);
    if (!dms) {
      if (i <= duc)
        ad(ans, (duc - i + 1) % mo * dlt % mo);
      else
        break;
    } else {
      if (i <= duc) ad(ans, (duc - i + 1) % mo * dlt % mo);
      if (i <= duc + 1)
        for (int j = 0; j < dms; j++) ad(ans, f[scnt[val[j + 1]]]);
      if (i > duc + 1) break;
    }
  }
  printf("%lld\n", ans);
  return 0;
}

### Prompt

Create a solution in CPP for the following problem:

While Duff was resting in the beach, she accidentally found a strange array b0, b1, ..., bl - 1 consisting of l positive integers. This array was strange because it was extremely long, but there was another (maybe shorter) array, a0, ..., an - 1 that b can be build from a with formula: bi = ai mod n where a mod b denoted the remainder of dividing a by b.

<image>

Duff is so curious, she wants to know the number of subsequences of b like bi1, bi2, ..., bix (0 ≤ i1 < i2 < ... < ix < l), such that: 

  * 1 ≤ x ≤ k
  * For each 1 ≤ j ≤ x - 1, <image>
  * For each 1 ≤ j ≤ x - 1, bij ≤ bij + 1. i.e this subsequence is non-decreasing. 



Since this number can be very large, she want to know it modulo 109 + 7.

Duff is not a programmer, and Malek is unavailable at the moment. So she asked for your help. Please tell her this number.

Input

The first line of input contains three integers, n, l and k (1 ≤ n, k, n × k ≤ 106 and 1 ≤ l ≤ 1018).

The second line contains n space separated integers, a0, a1, ..., an - 1 (1 ≤ ai ≤ 109 for each 0 ≤ i ≤ n - 1). 

Output

Print the answer modulo 1 000 000 007 in one line.

Examples

Input

3 5 3
5 9 1


Output

10


Input

5 10 3
1 2 3 4 5


Output

25

Note

In the first sample case, <image>. So all such sequences are: <image>, <image>, <image>, <image>, <image>, <image>, <image>, <image>, <image> and <image>.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
template <typename T1, typename T2>
inline T1 max(T1 a, T2 b) {
  return a < b ? b : a;
}
template <typename T1, typename T2>
inline T1 min(T1 a, T2 b) {
  return a < b ? a : b;
}
namespace ae86 {
const int bufl = 1 << 15;
char buf[bufl], *s = buf, *t = buf;
inline int fetch() {
  if (s == t) {
    t = (s = buf) + fread(buf, 1, bufl, stdin);
    if (s == t) return EOF;
  }
  return *s++;
}
inline long long ty() {
  long long a = 0;
  int b = 1, c = fetch();
  while (!isdigit(c)) b ^= c == '-', c = fetch();
  while (isdigit(c)) a = a * 10 + c - 48, c = fetch();
  return b ? a : -a;
}
}  // namespace ae86
using ae86::ty;
const int _ = 1000007, mo = 1000000007;
template <typename T1, typename T2>
inline T1 ad(T1 &a, T2 b) {
  return a = a + b >= mo ? a + b - mo : a + b;
}
template <typename T1, typename T2>
inline T1 dl(T1 &a, T2 b) {
  return a = a >= b ? a - b : a - b + mo;
}
inline long long powa(long long a, long long t) {
  long long b = 1;
  while (t) {
    if (t & 1) b = b * a % mo;
    a = a * a % mo, t >>= 1;
  }
  return b;
}
inline long long inva(long long a) { return powa(a, mo - 2); }
int n, tar, val[_], pval[_], ha[_] = {0}, lha = 0, cnt[_] = {0}, scnt[_] = {0},
                             loca[_] = {0};
long long len, f[_] = {0}, sf[_] = {0};
int main() {
  n = ty(), len = ty(), tar = ty();
  for (int i = 1; i <= n; i++) val[i] = ty(), ha[++lha] = val[i];
  sort(ha + 1, ha + lha + 1), lha = unique(ha + 1, ha + lha + 1) - ha - 1;
  for (int i = 1; i <= n; i++)
    val[i] = pval[i] = lower_bound(ha + 1, ha + lha + 1, val[i]) - ha,
    cnt[val[i]]++;
  for (int i = 1; i <= lha; i++) scnt[i] = scnt[i - 1] + cnt[i];
  sort(pval + 1, pval + n + 1);
  long long ans = len % mo, duc = len / n, dms = len % n;
  for (int i = 1; i <= n; i++) f[i] = 1;
  for (int i = 2; i <= tar; i++) {
    for (int j = 1; j <= n; j++) sf[j] = (sf[j - 1] + f[j]) % mo;
    long long dlt = 0;
    for (int j = 1; j <= n; j++) f[j] = sf[scnt[pval[j]]], ad(dlt, f[j]);
    if (!dms) {
      if (i <= duc)
        ad(ans, (duc - i + 1) % mo * dlt % mo);
      else
        break;
    } else {
      if (i <= duc) ad(ans, (duc - i + 1) % mo * dlt % mo);
      if (i <= duc + 1)
        for (int j = 0; j < dms; j++) ad(ans, f[scnt[val[j + 1]]]);
      if (i > duc + 1) break;
    }
  }
  printf("%lld\n", ans);
  return 0;
}

```