#include<bits/stdc++.h>
#define rep(i, n) for(int i = 0; i < (n); ++i)
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef vector<int> vi;
typedef pair<int, int> pii;
typedef pair<ll, int> pli;
typedef array<int, 3> ai3;
const int inf = 0x3f3f3f3f;
const ll INF = 0x3f3f3f3f3f3f3f3fll;
const int Mod = 1e9 + 7;
inline int sign(int a){ return (a&1) ? (Mod-1) : 1; }
inline void uadd(int &a, int b){ a += b-Mod; a += (a>>31) & Mod; }
inline void usub(int &a, int b){ a -= b, a += (a>>31) & Mod; }
inline void umul(int &a, int b){ a = (int)(1ll * a * b % Mod); }
inline int add(int a, int b){ a += b-Mod; a += (a>>31) & Mod; return a; }
inline int sub(int a, int b){ a -= b, a += (a>>31) & Mod; return a; }
inline int mul(int a, int b){ a = (int)(1ll * a * b % Mod); return a; }
int qpow(int b, ll p){ int ret = 1; while(p){ if(p&1) umul(ret, b); umul(b, b), p >>= 1; } return ret; }
const int fN = 4040;
int fact[fN], invfact[fN], pw2[fN], invpw2[fN], inv[fN];
void initfact(int n){
	pw2[0] = 1; for(int i = 1; i <= n; ++i) pw2[i] = mul(pw2[i-1], 2);
	invpw2[0] = 1; for(int i = 1; i <= n; ++i) invpw2[i] = mul(invpw2[i-1], (Mod+1) / 2);
	fact[0] = 1; for(int i = 1; i <= n; ++i) fact[i] = mul(fact[i-1], i);
	invfact[n] = qpow(fact[n], Mod-2); for(int i = n; i > 0; --i) invfact[i-1] = mul(invfact[i], i);
	for(int i = 1; i <= n; ++i) inv[i] = mul(invfact[i], fact[i-1]);
}
int binom(int n, int m){ return (m < 0 || m > n) ? 0 : mul(fact[n], mul(invfact[m], invfact[n-m])); }
const double pi = acos(-1);
inline void chmax(int &a, int b){ (b>a) ? (a=b) : 0; }
inline void chmin(int &a, int b){ (b<a) ? (a=b) : 0; }

int n;
string s;

string cans[] = {"ABCD", "BCDA", "CDAB", "DABC"};

//string st[2020];

int f[4][505], g3[505], g4[505], dp[2020][505];

inline bool ok(string tmp){
	return tmp == cans[0] || tmp == cans[1] || tmp == cans[2] || tmp == cans[3];
}

int main(){
	//freopen("a.in", "r", stdin);
	ios::sync_with_stdio(false);
	cin.tie(0);

	initfact(4004);

	cin >> s;
	n = (int)s.size();

	int m = 0, upd = 1;
	while(upd){
		upd = 0;
		rep(i, n - 3){
			if(ok(s.substr(i, 4))){
				//cout << i << endl;
				++upd, ++m;
				string tmp = s.substr(0, i) + s.substr(i + 4, n - i - 4);
				//cout << tmp << endl;
				s = tmp;
				n -= 4;
				break;
			}
		}
	}

	f[0][0] = f[1][0] = f[2][0] = f[3][0] = 1, g3[0] = g4[0] = 1;
	for(int i = 1; i <= m; ++i){
		rep(j, i){
			uadd(g3[i], mul(g3[j], mul(3, f[3][i-j-1])));
			uadd(g4[i], mul(g4[j], mul(4, f[3][i-j-1])));
		}
		rep(tp, 3){
			rep(j, i+1)
				uadd(f[tp + 1][i], mul(f[tp][i - j], g3[j]));
		}
		//cout << i << ": " << f[0][i] << " " << f[1][i] << " " << f[2][i] << " " << f[3][i] << "  " << g[i] << "\n";
	}

	dp[0][0] = 1;
	rep(i, n+1){
		rep(j, m+1){
			rep(k, m-j+1){
				//cout << i << " " << j << " " << k << ": " << dp[i][j] << " " << g3[k] << " " << g4[k] << "\n";
				uadd(dp[i+1][j+k], mul(dp[i][j], (i == n) ? g4[k] : g3[k]));
			}
		}
	}
	cout << dp[n+1][m] << "\n";

	return 0;
}