F

In the first test case,

$P(0) = F_2^{F_2} = 2^2 = 4\pmod {10000019}$, as $MEX(2, 2) = 0$,

$P(1) = F_1^{F_1} = 1^1 = 1\pmod {10000019}$, as $MEX(1, 1) = 1$, and

$P(2) = F_1^{F_2} = 1^2 = 1\pmod {10000019}$, as $MEX(1, 2) = 2$.

In the second test case,

$P(0) = F_1^{F_1} \cdot F_1^{F_2} \cdot F_1^{F_3} \cdot F_2^{F_2} \cdot F_2^{F_3} \cdot F_3^{F_3} = 1^1 \cdot 1^2 \cdot 1^3 \cdot 2^2 \cdot 2^3 \cdot 3^3 = 864\pmod {10000019}$, as $MEX(1, 1) = MEX(1, 2) = MEX(1, 3) = MEX(2, 2) = MEX(2, 3) = MEX(3, 3) = 0$,

$P(1) = 1\pmod {10000019}$,

$P(2) = 1\pmod {10000019}$, and

$P(3) = 1\pmod {10000019}$.

In the third test case,

$P(0) = F_2^{F_2} \cdot F_4^{F_4} \cdot F_4^{F_5} \cdot F_5^{F_5} = 2^2 \cdot 5^5 \cdot 5^8 \cdot 8^8 = 2295735\pmod {10000019}$, as $MEX(2, 2) = MEX(4, 4) = MEX(4, 5) = MEX(5, 5) = 0$,

$P(1) = F_1^{F_1} \cdot F_1^{F_2} \cdot F_1^{F_3} \cdot F_2^{F_3} \cdot F_3^{F_3} = 1^1 \cdot 1^2 \cdot 1^3 \cdot 2^3 \cdot 3^3 = 216\pmod {10000019}$, as $MEX(1, 1) = MEX(1, 2) = MEX(1, 3) = MEX(2, 3) = MEX(3, 3) = 1$,

$P(2) = F_1^{F_4} \cdot F_2^{F_4} \cdot F_3^{F_4} = 1^5 \cdot 2^5 \cdot 3^5 = 7776\pmod {10000019}$, as $MEX(1, 4) = MEX(2, 4) = MEX(3, 4) = 2$,

$P(3) = F_3^{F_5} = 3^8 = 6561\pmod {10000019}$, as $MEX(3, 5) = 3$,

$P(4) = F_1^{F_5} \cdot F_2^{F_5} = 1^8 \cdot 2^8 = 256\pmod {10000019}$, as $MEX(1, 5) = MEX(2, 5) = 4$, and

$P(5) = 1\pmod {10000019}$.