Problem when using interrupt INT0 and INT1

I am using interrupt INT0 and INT1 for two different sensors. They are both for counting detected object (see the code).

But I cant use them at the same time and I don't know what !'m doing wrong. Dig.pin 3 will change scount1 even though it is only supposed to change scount2.

If I put dig.pin 2 as input and dig.pin 3 as output (or the opposite) that counter works. If I remove the interrupt for dig.pin3 but keep it as input, it still mess up the counting... It is like the pins are connected.

Is it wrong to use both INT0 and INT1 at the same time? Please help! Thanx.

#include <avr/interrupt.h>

volatile int scount1=0;
volatile int scount2=0;
volatile unsigned long objectDetectedMillis;

static unsigned long timeoutMills = 120000;
 
void setup()
{
  Serial.begin(9600);
  pinMode(13, OUTPUT);     // Pin 13 is output to which an LED is connected
  digitalWrite(13, LOW);   // Make pin 13 low, switch LED off
  pinMode(2, INPUT);	   // Pin 2 is input to which a switch is connected = INT0
  pinMode(3, INPUT);	   // Pin 3 is input to which a switch is connected = INT1
  
  attachInterrupt(0, count1, FALLING);
  
  attachInterrupt(1, count2, FALLING);
}

void loop() {
  
  if (millis() - objectDetectedMillis >= timeoutMills) {
     digitalWrite(13, LOW);   // Make pin 13 low
     scount1 = 0;
     scount2 = 0;
  }
  
     Serial.println(scount1);
     Serial.println(scount2);
    
}  


void count1(){              // Interrupt service routine
  scount1 = scount1 + 1;
  digitalWrite(13, HIGH);
  
  objectDetectedMillis = millis();
 
}

void count2(){              // Interrupt service routine
   scount2 = scount2 + 1;
   objectDetectedMillis = millis();
 
   if (scount1 == scount2){
     digitalWrite(13, LOW);   // Make pin 13 low
     scount1 = 0;
     scount2 = 0;
     }

}

I see you're not using INPUT_PULLUP in your code so I assume the inputs are pulled up externally, since you're using FALLING interrupt.

That was the problem!!! Thank you!