#include <bits/stdc++.h>
using namespace std;
int t, n, a, b, len;
string s;
class fastio {
public:
    fastio() {
        ios_base::sync_with_stdio(false);
        cout.tie(nullptr);
        cin.tie(nullptr);
    }
} __fastio;
int f(int p, int i) {
    //Base Case
    if(p == 1 && i == len-1) return (a+b);
    else if(p == 2 && i == len-1) return ((2*a)+b);
    /*
      when there is crossroad (i.e 1) then x should be h = 2 and x + 1 could be h = 1 or 2 it depends
      on the value infront, for no crossroad (i.e 0) 'x' height could be either 1 or 2 depends on the
      optimal answer
     */
    if(s[i] == '0' && p == 1) {
        /*
           if the infront value is 0 and past value of pillar (here p in f(p)) is 1, then there are two
           cases either front pillar could be h = 1 or 2, if 1 then cost would be cost of
           pillar + gasline that is --> a + b
           if h = 2 then cost would be 2 * unit pillar + 2 * unit gasline
           that is --> 2(a+b)
         */
        return min(f(2,i+1)+2*(a+b), f(1,i+1)+a+b);
    }
    else if(s[i] == '1' && p == 1) {
        /*
          if infront value is 1, and past pillar is h = 1 then cost is 2 * pillar cost + gasline
         */
        return f(1,i+1)+2*(a+b);
    }
    else if(s[i] == '0' && p == 2) {
        /*
          if the past pillar height is 2 and the infront value is 0, then h = 1 or 2, if 1 then cost
          would be 2a + b else 2b + a
         */
        return min(f(1,i+1)+(2*a)+b, f(2,i+1)+(2*b)+a);
    }
    else if(s[i] == 1 && p == 2) {
        /*
          if the past pillar height is 2 and the infront value is 1 then h will be 2 and cost will be
          2b + a
         */
        return f(2,i+1)+(2*b)+a;
    }
}
void solve() {
    cin >> n >> a >> b;
    cin >> s;
    len = s.length();
    cout << b + f(1,1) << endl;
}
int main() {
  __fastio;
  cin >> t;
  while(t--) {
      solve();
  }
  return 0;
}